| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the function \(\frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}\), we will use the quotient rule. The quotient rule states that if we have a function \( \frac{u}{v} \), its derivative is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] Here, let \( u = \log(10) \cos \left(\sqrt{x+8}\right) \) and \( v = \log(5x) \). First, we need to find the derivatives \( u' \) and \( v' \). ### Step 1: Find \( u' \) \[ u = \log(10) \cos \left(\sqrt{x+8}\right) \] Since \(\log(10)\) is a constant, we only need to differentiate \(\cos \left(\sqrt{x+8}\right)\): \[ \frac{d}{dx} \cos \left(\sqrt{x+8}\right) = -\sin \left(\sqrt{x+8}\right) \cdot \frac{d}{dx} \left(\sqrt{x+8}\right) \] Next, we differentiate \(\sqrt{x+8}\): \[ \frac{d}{dx} \left(\sqrt{x+8}\right) = \frac{1}{2\sqrt{x+8}} \cdot \frac{d}{dx} (x+8) = \frac{1}{2\sqrt{x+8}} \] So, \[ \frac{d}{dx} \cos \left(\sqrt{x+8}\right) = -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \] Thus, \[ u' = \log(10) \left( -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \right) = -\log(10) \cdot \frac{\sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \] ### Step 2: Find \( v' \) \[ v = \log(5x) \] Using the chain rule, \[ \frac{d}{dx} \log(5x) = \frac{1}{5x} \cdot \frac{d}{dx} (5x) = \frac{1}{5x} \cdot 5 = \frac{1}{x} \] So, \[ v' = \frac{1}{x} \] ### Step 3: Apply the Quotient Rule Now, we apply the quotient rule: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] Substitute \( u \), \( u' \), \( v \), and \( v' \): \[ \left( \frac{\log(10) \cos \left(\sqrt{x+8}\right)}{\log(5x)} \right)' = \frac{\left( -\log(10) \cdot \frac{\sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \right) \log(5x) - \log(10) \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{\left( \log(5x) \right)^2} \] Simplify the numerator: \[ = \frac{-\log(10) \cdot \frac{\sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \cdot \log(5x) - \log(10) \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{\left( \log(5x) \right)^2} \] Factor out \(-\log(10)\): \[ = \frac{-\log(10) \left( \frac{\sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \cdot \log(5x) + \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x} \right)}{\left( \log(5x) \right)^2} \] Thus, the final answer is: \[ \boxed{\frac{-\log(10) \left( \frac{\sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \cdot \log(5x) + \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x} \right)}{\left( \log(5x) \right)^2}} \] |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| deepseek-r1 | To differentiate the function \(\frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}\), we use the quotient rule. Assuming \(\log\) denotes the natural logarithm (base \(e\)), the function becomes \(\frac{\ln(10) \cos(\sqrt{x + 8})}{\ln(5x)}\). 1. **Identify the numerator and denominator:** - Numerator \(u = \ln(10) \cos(\sqrt{x + 8})\) - Denominator \(v = \ln(5x)\) 2. **Compute the derivatives of \(u\) and \(v\):** - Derivative of \(u\): \[ u' = \ln(10) \cdot \left(-\sin(\sqrt{x + 8}) \cdot \frac{1}{2\sqrt{x + 8}}\right) = -\frac{\ln(10) \sin(\sqrt{x + 8})}{2\sqrt{x + 8}} \] - Derivative of \(v\): \[ v' = \frac{1}{x} \] 3. **Apply the quotient rule \(\frac{u'v - uv'}{v^2}\):** \[ \frac{d}{dx}\left(\frac{\ln(10) \cos(\sqrt{x + 8})}{\ln(5x)}\right) = \frac{-\frac{\ln(10) \sin(\sqrt{x + 8}) \ln(5x)}{2\sqrt{x + 8}} - \frac{\ln(10) \cos(\sqrt{x + 8})}{x}}{(\ln(5x))^2} \] 4. **Factor out common terms and simplify:** \[ -\frac{\ln(10)}{(\ln(5x))^2} \left( \frac{\sin(\sqrt{x + 8}) \ln(5x)}{2\sqrt{x + 8}} + \frac{\cos(\sqrt{x + 8})}{x} \right) \] Thus, the final answer is: \[ \boxed{ -\frac{\ln(10) \left( \frac{\sin\left(\sqrt{x + 8}\right) \ln(5x)}{2\sqrt{x + 8}} + \frac{\cos\left(\sqrt{x + 8}\right)}{x} \right)}{\left(\ln(5x)\right)^2} } \] |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| gpto1 | To find the derivative of the function: \[ f(x) = \frac{\log(10) \cos \left(\sqrt{x+8}\right)}{\log(5 x)} \] Since \(\log_{10}(10) = 1\), the function simplifies to: \[ f(x) = \frac{ \cos \left( \sqrt{x+8} \right ) }{ \log_{10} (5 x)} \] Now, set: 1. \( u(x) = \cos \left( \sqrt{x+8} \right ) \) 2. \( v(x) = \log_{10}(5 x) \) Compute \(u'(x)\) and \(v'(x)\): For \(u(x)\): \[ u'(x) = -\sin \left( \sqrt{x+8} \right ) \cdot \frac{1}{2 \sqrt{x+8}} \] For \(v(x)\): \[ v'(x) = \frac{1}{ x \ln 10 } \] Now apply the quotient rule: \[ f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2} \] Plugging in the expressions for \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\): \[ f'(x) = \frac{ -\sin \left( \sqrt{x+8} \right ) \cdot \frac{1}{2 \sqrt{x+8}} \cdot \log_{10}(5 x ) - \cos \left( \sqrt{x+8} \right ) \cdot \frac{1}{ x \ln 10 } }{ \left[ \log_{10} (5 x ) \right ]^2 } \] Simplify the numerator: \[ N(x) = - \left( \frac{ \sin \left( \sqrt{x+8} \right ) \log_{10}(5 x ) }{2 \sqrt{x+8} } + \frac{ \cos \left( \sqrt{x+8} \right ) }{ x \ln 10 } \right ) \] Find a common denominator in the numerator (\(2 x \sqrt{x+8} \ln 10\)) and combine terms: \[ N(x) = - \left( \frac{ x \ln 10 \sin \left( \sqrt{x+8} \right ) \log_{10}(5 x ) + 2 \sqrt{x+8} \cos \left( \sqrt{x+8} \right ) }{ 2 x \sqrt{x+8} \ln 10 } \right ) \] Now, write the derivative as: \[ f'(x) = - \frac{ x \ln 10 \sin \left( \sqrt{x+8} \right ) \log_{10}(5 x ) + 2 \sqrt{x+8} \cos \left( \sqrt{x+8} \right ) }{ 2 x \sqrt{x+8} \ln 10 \left[ \log_{10}(5 x ) \right ]^2 } \] Thus, the derivative is: \(\boxed{\, -\,\dfrac{x\,\ln10\ \sin\bigl(\sqrt{x+8}\bigr)\,\log_{10}(5x)\,+\,2\sqrt{x+8}\ \cos\bigl(\sqrt{x+8}\bigr)}{2x\,\sqrt{x+8}\ \ln10\ \left[\log_{10}(5x)\right]^{\,2}}\,}\) |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| gpt-4o-2024-08-06 | To differentiate the function \( y = \frac{\log(10) \cos(\sqrt{x+8})}{\log(5x)} \), we will use the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then \[ y' = \frac{u'v - uv'}{v^2} \] For our function, let \[ u = \log(10) \cos(\sqrt{x+8}) \] \[ v = \log(5x) \] First, we differentiate \( u \): \[ u = \log(10) \cos(\sqrt{x+8}) \] Since \(\log(10)\) is a constant, we have: \[ u' = \log(10) \cdot \frac{d}{dx} \left( \cos(\sqrt{x+8}) \right) \] Using the chain rule, the derivative of \(\cos(\sqrt{x+8})\) is: \[ \frac{d}{dx} \left( \cos(\sqrt{x+8}) \right) = -\sin(\sqrt{x+8}) \cdot \frac{d}{dx}(\sqrt{x+8}) \] The derivative of \(\sqrt{x+8}\) is: \[ \frac{d}{dx}(\sqrt{x+8}) = \frac{1}{2\sqrt{x+8}} \] Thus, \[ u' = \log(10) \left( -\sin(\sqrt{x+8}) \cdot \frac{1}{2\sqrt{x+8}} \right) = -\frac{\log(10) \sin(\sqrt{x+8})}{2\sqrt{x+8}} \] Next, we differentiate \( v \): \[ v = \log(5x) \] Using the chain rule, the derivative is: \[ v' = \frac{1}{5x} \cdot 5 = \frac{1}{x} \] Now, apply the quotient rule: \[ y' = \frac{u'v - uv'}{v^2} \] Substitute \( u' \), \( u \), \( v' \), and \( v \): \[ y' = \frac{\left(-\frac{\log(10) \sin(\sqrt{x+8})}{2\sqrt{x+8}}\right) \log(5x) - \log(10) \cos(\sqrt{x+8}) \cdot \frac{1}{x}}{(\log(5x))^2} \] Simplify the expression: \[ y' = \frac{-\frac{\log(10) \sin(\sqrt{x+8}) \log(5x)}{2\sqrt{x+8}} - \frac{\log(10) \cos(\sqrt{x+8})}{x}}{(\log(5x))^2} \] Combine the terms in the numerator: \[ y' = \frac{-\log(10) \left( \frac{\sin(\sqrt{x+8}) \log(5x)}{2\sqrt{x+8}} + \frac{\cos(\sqrt{x+8})}{x} \right)}{(\log(5x))^2} \] Thus, the derivative of the function is: \[ \boxed{\frac{-\log(10) \left( \frac{\sin(\sqrt{x+8}) \log(5x)}{2\sqrt{x+8}} + \frac{\cos(\sqrt{x+8})}{x} \right)}{(\log(5x))^2}} \] |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| DeepSeek-Coder-V2-Lite-Instruct | To differentiate the function \(\frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}\), we will use the quotient rule. The quotient rule states that if we have a function \( \frac{u(x)}{v(x)} \), its derivative is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] Here, \( u(x) = \log (10) \cos \left(\sqrt{x+8}\right) \) and \( v(x) = \log (5x) \). First, we need to find the derivatives of \( u(x) \) and \( v(x) \). ### Step 1: Differentiate \( u(x) \) The derivative of \( \log (10) \) is 0 because \( \log (10) \) is a constant. Therefore, we only need to differentiate \( \cos \left(\sqrt{x+8}\right) \). Let \( w = \sqrt{x+8} \). Then, \( \cos \left(\sqrt{x+8}\right) = \cos(w) \). The derivative of \( \cos(w) \) with respect to \( w \) is \( -\sin(w) \). Now, we need to find the derivative of \( w \) with respect to \( x \): \[ w = \sqrt{x+8} \implies \frac{dw}{dx} = \frac{1}{2\sqrt{x+8}} \] Thus, the derivative of \( \cos \left(\sqrt{x+8}\right) \) with respect to \( x \) is: \[ \frac{d}{dx} \cos \left(\sqrt{x+8}\right) = -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \] So, \( u'(x) = \log (10) \cdot \left( -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \right) \). ### Step 2: Differentiate \( v(x) \) The derivative of \( \log (5x) \) can be found using the chain rule. Let \( u = 5x \). Then, \( \log (5x) = \log(u) \). The derivative of \( \log(u) \) with respect to \( u \) is \( \frac{1}{u} \). The derivative of \( u = 5x \) with respect to \( x \) is \( 5 \). Thus, the derivative of \( \log (5x) \) with respect to \( x \) is: \[ \frac{d}{dx} \log (5x) = \frac{1}{5x} \cdot 5 = \frac{1}{x} \] So, \( v'(x) = \frac{1}{x} \). ### Step 3: Apply the quotient rule Now, we apply the quotient rule: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ \left( \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5x)} \right)' = \frac{\left( \log (10) \cdot \left( -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \right) \right) \log (5x) - \log (10) \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{(\log (5x))^2} \] Simplify the expression: \[ = \frac{-\log (10) \sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \log (5x) - \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{x}}{(\log (5x))^2} \] \[ = \frac{-\log (10) \sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \log (5x) - \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{x}}{(\log (5x))^2} \] \[ = \frac{-\log (10) \sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \log (5x) - \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{x}}{(\log (5x))^2} \] This is a complex expression, and further simplification might be possible, but it is already quite involved. The final answer is: \[ \boxed{\frac{-\log (10) \sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \log (5x) - \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{x}}{(\log (5x))^2}} \] |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $\frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}$. ## Step 2: Recognize that the function is a quotient of two functions Since the function is a quotient, we will use the quotient rule for differentiation, which states that if $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{[h(x)]^2}$. ## Step 3: Identify g(x) and h(x) for the quotient rule Let $g(x) = \log (10) \cos \left(\sqrt{x+8}\right)$ and $h(x) = \log (5 x)$. ## Step 4: Differentiate g(x) To differentiate $g(x)$, we use the product rule since it is a product of two functions. The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. Here, $u(x) = \log (10)$, which is a constant, and $v(x) = \cos \left(\sqrt{x+8}\right)$. The derivative of a constant is 0, so $u'(x) = 0$. To find $v'(x)$, we use the chain rule because $v(x)$ is a composite function. The derivative of $\cos(x)$ is $-\sin(x)$, and the derivative of $\sqrt{x+8}$ is $\frac{1}{2\sqrt{x+8}}$. Therefore, $v'(x) = -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}}$. Thus, $g'(x) = 0 \cdot \cos \left(\sqrt{x+8}\right) + \log (10) \cdot \left(-\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}}\right)$, which simplifies to $g'(x) = -\frac{\log (10) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}}$. ## Step 5: Differentiate h(x) To differentiate $h(x) = \log (5 x)$, we use the chain rule. The derivative of $\log(x)$ is $\frac{1}{x}$, and the derivative of $5x$ is $5$. Therefore, $h'(x) = \frac{1}{5x} \cdot 5 = \frac{1}{x}$. ## Step 6: Apply the quotient rule Substitute $g(x)$, $h(x)$, $g'(x)$, and $h'(x)$ into the quotient rule formula: $f'(x) = \frac{\log (5 x) \cdot \left(-\frac{\log (10) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}}\right) - \log (10) \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{[\log (5 x)]^2}$. ## Step 7: Simplify the derivative Simplify the expression for $f'(x)$: $f'(x) = \frac{-\frac{\log (10) \log (5 x) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} - \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{x}}{[\log (5 x)]^2}$. ## Step 8: Further simplification This expression can be further simplified by finding a common denominator for the terms in the numerator and then combining them, but given the complexity and the instruction to follow the format to the letter, we proceed to present the final answer as is, acknowledging that additional algebraic simplification could be performed. The final answer is: $\boxed{\frac{-\frac{\log (10) \log (5 x) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} - \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{x}}{[\log (5 x)]^2}}$ |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $\frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}$. This is a quotient of two functions, involving logarithmic and trigonometric functions. ## Step 2: Apply the quotient rule of differentiation The quotient rule states that if we have a function of the form $f(x) = \frac{g(x)}{h(x)}$, then the derivative $f'(x)$ is given by $f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{[h(x)]^2}$. Here, $g(x) = \log (10) \cos \left(\sqrt{x+8}\right)$ and $h(x) = \log (5 x)$. ## Step 3: Differentiate the numerator $g(x)$ To differentiate $g(x)$, we'll apply the product rule and the composite function rule. The derivative of $\cos \left(\sqrt{x+8}\right)$ is $-\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}}$. Since $\log (10)$ is a constant, it can be treated as a coefficient. Thus, $g'(x) = \log (10) \cdot \left(-\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}}\right)$. ## Step 4: Differentiate the denominator $h(x)$ The derivative of $\log (5 x)$ is $\frac{1}{5x} \cdot 5 = \frac{1}{x}$, using the chain rule and the fact that the derivative of $\log (u)$ is $\frac{1}{u}$. ## Step 5: Substitute $g(x)$, $h(x)$, $g'(x)$, and $h'(x)$ into the quotient rule formula Substituting the expressions for $g(x)$, $h(x)$, $g'(x)$, and $h'(x)$ into the quotient rule formula, we get $f'(x) = \frac{\log (5 x) \cdot \left(\log (10) \cdot \left(-\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}}\right)\right) - \log (10) \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{[\log (5 x)]^2}$. ## Step 6: Simplify the expression for $f'(x)$ Simplifying the expression for $f'(x)$, we get $f'(x) = \frac{\log (5 x) \cdot \left(-\log (10) \cdot \sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}}\right) - \log (10) \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{[\log (5 x)]^2}$. ## Step 7: Factor out common terms and simplify further Factoring out common terms and simplifying further, we get $f'(x) = \frac{-\log (10)}{[\log (5 x)]^2} \cdot \left(\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} + \frac{\cos \left(\sqrt{x+8}\right)}{x}\right)$. ## Step 8: Write the final answer in the required format The final answer is: $\boxed{\frac{-\log (10)}{[\log (5 x)]^2} \cdot \left(\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} + \frac{\cos \left(\sqrt{x+8}\right)}{x}\right)}$ |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| Qwen2-72B-Instruct | To differentiate the given function, we will use the quotient rule and chain rule as necessary. The function is: \[f(x) = \frac{\log(10) \cos\left(\sqrt{x+8}\right)}{\log(5x)}\] Let's denote the numerator by \(u\) and the denominator by \(v\), so that: \[u = \log(10) \cos\left(\sqrt{x+8}\right)\] \[v = \log(5x)\] First, let's find the derivatives of \(u\) and \(v\): For \(u\): \[ u' = -\log(10) \sin\left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} = -\frac{\log(10)}{2\sqrt{x+8}} \sin\left(\sqrt{x+8}\right) \] For \(v\): \[ v' = \frac{1}{5x} \cdot 5 = \frac{1}{x} \] Now, using the quotient rule for differentiation, which states that if \(f(x) = \frac{u}{v}\), then \(f'(x) = \frac{u'v - uv'}{v^2}\), we have: \[ f'(x) = \frac{-\frac{\log(10)}{2\sqrt{x+8}} \sin\left(\sqrt{x+8}\right) \cdot \log(5x) - \log(10) \cos\left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{\log^2(5x)} \] Simplifying this expression: \[ f'(x) = \frac{-\log(10)\left[\frac{\sin\left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \log(5x) + \frac{\cos\left(\sqrt{x+8}\right)}{x}\right]}{\log^2(5x)} \] This is the derivative of the given function. \[ \boxed{f'(x) = \frac{-\log(10)\left[\frac{\sin\left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \log(5x) + \frac{\cos\left(\sqrt{x+8}\right)}{x}\right]}{\log^2(5x)}} \] |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| baichuan4 | To differentiate the function $f(x) = \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}$, we will use the quotient rule, the chain rule, and the product rule as needed. First, let's denote the numerator and denominator separately: - Numerator: $u(x) = \log (10) \cos \left(\sqrt{x+8}\right)$ - Denominator: $v(x) = \log (5 x)$ Using the quotient rule, we have: $$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$ Now, we need to find $u'(x)$ and $v'(x)$. For $u(x)$, we have a product of constants and a function, so we can treat $\log(10)$ as a constant and only differentiate $\cos\left(\sqrt{x+8}\right)$. Using the chain rule, we get: $$ u'(x) = \log(10) \cdot \left(-\sin\left(\sqrt{x+8}\right)\right) \cdot \frac{1}{2\sqrt{x+8}} \cdot \frac{d}{dx}(x+8) $$ $$ u'(x) = -\frac{\log(10) \sin\left(\sqrt{x+8}\right)}{2\sqrt{x+8}} $$ For $v(x)$, we use the chain rule to differentiate $\log(5x)$: $$ v'(x) = \frac{1}{5x} \cdot 5 $$ $$ v'(x) = \frac{1}{x} $$ Now, we can substitute $u'(x)$ and $v'(x)$ into the quotient rule formula: $$ f'(x) = \frac{-\frac{\log(10) \sin\left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \cdot \log(5x) - \log(10) \cos\left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{[\log(5x)]^2} $$ Simplifying, we get: $$ f'(x) = \frac{-\log(10) \sin\left(\sqrt{x+8}\right) \log(5x)}{2\sqrt{x+8}[\log(5x)]^2} - \frac{\log(10) \cos\left(\sqrt{x+8}\right)}{x[\log(5x)]^2} $$ Combining the terms, we have: $$ f'(x) = \frac{-\log(10) \left(\sin\left(\sqrt{x+8}\right) \log(5x) + 2\cos\left(\sqrt{x+8}\right)\sqrt{x+8}\right)}{2x\sqrt{x+8}[\log(5x)]^2} $$ Finally, we can put the answer in a box: $$ \boxed{f'(x) = \frac{-\log(10) \left(\sin\left(\sqrt{x+8}\right) \log(5x) + 2\cos\left(\sqrt{x+8}\right)\sqrt{x+8}\right)}{2x\sqrt{x+8}[\log(5x)]^2}} $$ |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| doubao | Let $y = \frac{\log (10) \cos(\sqrt{x+8})}{\log (5x)}$. Using the quotient rule: \[ \begin{align*} y'&=\frac{[\log(10)\cos(\sqrt{x+8})]'\cdot\log(5x)-\log(10)\cos(\sqrt{x+8})\cdot[\log(5x)]'}{(\log(5x))^2}\\ &=\frac{[\log(10)\times(-\sin(\sqrt{x+8}))\times\frac{1}{2\sqrt{x+8}}]\cdot\log(5x)-\log(10)\cos(\sqrt{x+8})\cdot\frac{5}{5x}}{(\log(5x))^2}\\ &=\frac{-\frac{\log(10)\sin(\sqrt{x+8})}{2\sqrt{x+8}}\log(5x)-\frac{\log(10)\cos(\sqrt{x+8})}{x}}{(\log(5x))^2}\\ \end{align*} \] Putting it in the box: $\boxed{\frac{-\frac{\log(10)\sin(\sqrt{x+8})}{2\sqrt{x+8}}\log(5x)-\frac{\log(10)\cos(\sqrt{x+8})}{x}}{(\log(5x))^2}}$. |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| gpto1 | To find the derivative of the function: \[ f(x) = \frac{\log(10) \cos \left(\sqrt{x+8}\right)}{\log(5 x)} \] Since \(\log_{10}(10) = 1\), the function simplifies to: \[ f(x) = \frac{ \cos \left( \sqrt{x+8} \right ) }{ \log_{10} (5 x)} \] Now, set: 1. \( u(x) = \cos \left( \sqrt{x+8} \right ) \) 2. \( v(x) = \log_{10}(5 x) \) Compute \(u'(x)\) and \(v'(x)\): For \(u(x)\): \[ u'(x) = -\sin \left( \sqrt{x+8} \right ) \cdot \frac{1}{2 \sqrt{x+8}} \] For \(v(x)\): \[ v'(x) = \frac{1}{ x \ln 10 } \] Now apply the quotient rule: \[ f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2} \] Plugging in the expressions for \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\): \[ f'(x) = \frac{ -\sin \left( \sqrt{x+8} \right ) \cdot \frac{1}{2 \sqrt{x+8}} \cdot \log_{10}(5 x ) - \cos \left( \sqrt{x+8} \right ) \cdot \frac{1}{ x \ln 10 } }{ \left[ \log_{10} (5 x ) \right ]^2 } \] Simplify the numerator: \[ N(x) = - \left( \frac{ \sin \left( \sqrt{x+8} \right ) \log_{10}(5 x ) }{2 \sqrt{x+8} } + \frac{ \cos \left( \sqrt{x+8} \right ) }{ x \ln 10 } \right ) \] Find a common denominator in the numerator (\(2 x \sqrt{x+8} \ln 10\)) and combine terms: \[ N(x) = - \left( \frac{ x \ln 10 \sin \left( \sqrt{x+8} \right ) \log_{10}(5 x ) + 2 \sqrt{x+8} \cos \left( \sqrt{x+8} \right ) }{ 2 x \sqrt{x+8} \ln 10 } \right ) \] Now, write the derivative as: \[ f'(x) = - \frac{ x \ln 10 \sin \left( \sqrt{x+8} \right ) \log_{10}(5 x ) + 2 \sqrt{x+8} \cos \left( \sqrt{x+8} \right ) }{ 2 x \sqrt{x+8} \ln 10 \left[ \log_{10}(5 x ) \right ]^2 } \] Thus, the derivative is: \(\boxed{\, -\,\dfrac{x\,\ln10\ \sin\bigl(\sqrt{x+8}\bigr)\,\log_{10}(5x)\,+\,2\sqrt{x+8}\ \cos\bigl(\sqrt{x+8}\bigr)}{2x\,\sqrt{x+8}\ \ln10\ \left[\log_{10}(5x)\right]^{\,2}}\,}\) |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| kimi | To differentiate the function \( f(x) = \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)} \), we will use the quotient rule. The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}. \] Here, \( u(x) = \log (10) \cos \left(\sqrt{x+8}\right) \) and \( v(x) = \log (5 x) \). First, we need to find the derivatives \( u'(x) \) and \( v'(x) \). 1. **Find \( u'(x) \):** \[ u(x) = \log (10) \cos \left(\sqrt{x+8}\right). \] Since \(\log (10)\) is a constant, we can factor it out: \[ u'(x) = \log (10) \cdot \frac{d}{dx} \left( \cos \left(\sqrt{x+8}\right) \right). \] Using the chain rule, we get: \[ \frac{d}{dx} \left( \cos \left(\sqrt{x+8}\right) \right) = -\sin \left(\sqrt{x+8}\right) \cdot \frac{d}{dx} \left( \sqrt{x+8} \right). \] The derivative of \(\sqrt{x+8}\) is: \[ \frac{d}{dx} \left( \sqrt{x+8} \right) = \frac{1}{2\sqrt{x+8}}. \] Therefore, \[ u'(x) = \log (10) \cdot \left( -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \right) = -\frac{\log (10) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}}. \] 2. **Find \( v'(x) \):** \[ v(x) = \log (5 x). \] Using the properties of logarithms, we can rewrite this as: \[ v(x) = \log (5) + \log (x). \] Since \(\log (5)\) is a constant, its derivative is zero: \[ v'(x) = \frac{d}{dx} (\log (x)) = \frac{1}{x}. \] Now, we apply the quotient rule: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}. \] Substituting \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the formula, we get: \[ f'(x) = \frac{\left( -\frac{\log (10) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \right) \log (5 x) - \left( \log (10) \cos \left(\sqrt{x+8}\right) \right) \left( \frac{1}{x} \right)}{[\log (5 x)]^2}. \] Simplifying the numerator: \[ f'(x) = \frac{-\frac{\log (10) \sin \left(\sqrt{x+8}\right) \log (5 x)}{2\sqrt{x+8}} - \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{x}}{[\log (5 x)]^2}. \] Factoring out \(-\log (10)\) from the numerator: \[ f'(x) = \frac{-\log (10) \left( \frac{\sin \left(\sqrt{x+8}\right) \log (5 x)}{2\sqrt{x+8}} + \frac{\cos \left(\sqrt{x+8}\right)}{x} \right)}{[\log (5 x)]^2}. \] Combining the terms in the parentheses: \[ f'(x) = \frac{-\log (10) \left( \frac{x \sin \left(\sqrt{x+8}\right |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| minimax | To differentiate the function $f(x) = \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}$, we will use the quotient rule. The quotient rule states that if you have a function $f(x) = \frac{g(x)}{h(x)}$, then its derivative $f'(x)$ is given by: $$ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} $$ Let's define $g(x)$ and $h(x)$ as follows: $$ g(x) = \log (10) \cos \left(\sqrt{x+8}\right) $$ $$ h(x) = \log (5 x) $$ First, we need to find the derivatives of $g(x)$ and $h(x)$. For $g(x)$, we have a product of two functions, so we'll use the product rule. Let $u(x) = \log (10)$ and $v(x) = \cos \left(\sqrt{x+8}\right)$. Since $\log (10)$ is a constant, its derivative |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| wenxinyiyan | Let $u(x)=\log (10) \cos \left(\sqrt{x+8}\right)$ and $v(x)=\log (5 x)$. Then, by the quotient rule, \begin{align*} \frac{d}{dx}\left(\frac{u}{v}\right)&=\frac{v \frac{d}{dx}(u)-u \frac{d}{dx}(v)}{v^2} \\ &=\frac{\log (5 x) \frac{d}{dx}\left(\log (10) \cos \left(\sqrt{x+8}\right)\right)-\log (10) \cos \left(\sqrt{x+8}\right) \frac{d}{dx}(\log (5 x))}{(\log (5 x))^2} \end{align*} Let's compute $\frac{d}{dx}\left(\log (10) \cos \left(\sqrt{x+8}\right)\right)$ and $\frac{d}{dx}(\log (5 x))$ separately. \begin{align*} \frac{d}{dx}\left(\log (10) \cos \left(\sqrt{x+8}\right)\right)&=\log (10) \frac{d}{dx}\left(\cos \left(\sqrt{x+8}\right)\right) \\ &=-\log (10) \sin \left(\sqrt{x+8}\right) \frac{d}{dx}\left(\sqrt{x+8}\right) \\ &=-\frac{\log (10)}{2 \sqrt{x+8}} \sin \left(\sqrt{x+8}\right) \end{align*} \begin{align*} \frac{d}{dx}(\log (5 x))&=\frac{1}{x} \end{align*} So, \begin{align*} \frac{d}{dx}\left(\frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)}\right)&=\frac{-\frac{\log (10)}{2 \sqrt{x+8}} \sin \left(\sqrt{x+8}\right) \log (5 x)-\frac{\log (10)}{x} \cos \left(\sqrt{x+8}\right)}{(\log (5 x))^2} \\ &=\boxed{\frac{-\log (10) \left(\frac{\log (5 x)}{2 \sqrt{x+8}} \sin \left(\sqrt{x+8}\right)+\frac{1}{x} \cos \left(\sqrt{x+8}\right)\right)}{(\log (5 x))^2}} \end{align*} |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| zhipu | To differentiate the function \( f(x) = \frac{\log (10) \cos \left(\sqrt{x+8}\right)}{\log (5 x)} \), we will use the quotient rule. The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] Here, let: \[ u(x) = \log(10) \cos \left(\sqrt{x+8}\right) \] \[ v(x) = \log(5x) \] First, we need to find \( u'(x) \) and \( v'(x) \). ### Finding \( u'(x) \): \[ u(x) = \log(10) \cos \left(\sqrt{x+8}\right) \] Since \(\log(10)\) is a constant, we focus on differentiating \(\cos \left(\sqrt{x+8}\right)\). We use the chain rule: \[ \frac{d}{dx} \cos \left(\sqrt{x+8}\right) = -\sin \left(\sqrt{x+8}\right) \cdot \frac{d}{dx} \sqrt{x+8} \] Next, we differentiate \(\sqrt{x+8}\): \[ \frac{d}{dx} \sqrt{x+8} = \frac{1}{2\sqrt{x+8}} \] Putting it all together: \[ u'(x) = \log(10) \left( -\sin \left(\sqrt{x+8}\right) \cdot \frac{1}{2\sqrt{x+8}} \right) = -\frac{\log(10) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \] ### Finding \( v'(x) \): \[ v(x) = \log(5x) \] Using the property of logarithms, \(\log(5x) = \log(5) + \log(x)\). Since \(\log(5)\) is a constant, we differentiate \(\log(x)\): \[ v'(x) = \frac{d}{dx} (\log(5) + \log(x)) = \frac{d}{dx} \log(x) = \frac{1}{x} \] ### Applying the Quotient Rule: Now, substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] \[ f'(x) = \frac{\left( -\frac{\log(10) \sin \left(\sqrt{x+8}\right)}{2\sqrt{x+8}} \right) \log(5x) - \log(10) \cos \left(\sqrt{x+8}\right) \cdot \frac{1}{x}}{[\log(5x)]^2} \] Simplify the numerator: \[ f'(x) = \frac{-\frac{\log(10) \sin \left(\sqrt{x+8}\right) \log(5x)}{2\sqrt{x+8}} - \frac{\log(10) \cos \left(\sqrt{x+8}\right)}{x}}{[\log(5x)]^2} \] Combine the terms in the numerator under a common denominator: \[ f'(x) = \frac{-\log(10) \sin \left(\sqrt{x+8}\right) \log(5x) / (2\sqrt{x+8}) - \log(10) \cos \left(\sqrt{x+8}\right) / x}{[\log(5x)]^2} \] \[ f'(x) = \frac{-\log(10) \left( \frac{\sin \left(\sqrt{x+8}\right) \log(5x)}{2\sqrt{x+8}} + \frac{\cos \left(\sqrt{x+8}\right)}{x} \right)}{[\log(5x)]^2} \] ### Final Answer: \[ \boxed{f'(x) = \frac{-\log(10) \left( \frac{\sin \left(\sqrt{x+8}\right) \log(5x)}{2\sqrt{x+8}} + \frac{\cos \left(\sqrt{x+8}\right)}{x} \right)}{[\log(5x)]^2}} \] |
\frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-23adbe6a-4cd3-9180-b74a-1c8239ebb003', 'request_id': '23adbe6a-4cd3-9180-b74a-1c8239ebb003'} | \frac{\log (10) \left(-\frac{2 \cos \left(\sqrt{x+8}\right)}{x}-\frac{\log (5 x) \sin \left(\sqrt{x+8}\right)}{\sqrt{x+8}}\right)}{2 \log ^2(5 x)} |