| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the function \(\sqrt{4-5x^3} \cot (8-4x)\), we will use the product rule and the chain rule. Let's denote the function as \(f(x) = \sqrt{4-5x^3} \cdot \cot(8-4x)\). First, we identify the two parts of the product: \[ u(x) = \sqrt{4-5x^3} \] \[ v(x) = \cot(8-4x) \] We need to find the derivatives of \(u(x)\) and \(v(x)\). ### Step 1: Differentiate \(u(x)\) \[ u(x) = (4-5x^3)^{1/2} \] Using the chain rule: \[ u'(x) = \frac{1}{2}(4-5x^3)^{-1/2} \cdot \frac{d}{dx}(4-5x^3) \] \[ \frac{d}{dx}(4-5x^3) = -15x^2 \] So, \[ u'(x) = \frac{1}{2}(4-5x^3)^{-1/2} \cdot (-15x^2) \] \[ u'(x) = -\frac{15x^2}{2\sqrt{4-5x^3}} \] ### Step 2: Differentiate \(v(x)\) \[ v(x) = \cot(8-4x) \] Using the chain rule: \[ v'(x) = -\csc^2(8-4x) \cdot \frac{d}{dx}(8-4x) \] \[ \frac{d}{dx}(8-4x) = -4 \] So, \[ v'(x) = -\csc^2(8-4x) \cdot (-4) \] \[ v'(x) = 4\csc^2(8-4x) \] ### Step 3: Apply the Product Rule \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substitute the derivatives we found: \[ f'(x) = \left(-\frac{15x^2}{2\sqrt{4-5x^3}}\right) \cot(8-4x) + \sqrt{4-5x^3} \cdot 4\csc^2(8-4x) \] ### Step 4: Simplify the Expression \[ f'(x) = -\frac{15x^2 \cot(8-4x)}{2\sqrt{4-5x^3}} + 4\sqrt{4-5x^3} \csc^2(8-4x) \] Thus, the final differentiated function is: \[ \boxed{-\frac{15x^2 \cot(8-4x)}{2\sqrt{4-5x^3}} + 4\sqrt{4-5x^3} \csc^2(8-4x)} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| deepseek-r1 | To differentiate the function \(\sqrt{4 - 5x^3} \cot(8 - 4x)\), we use the product rule. The product rule states that if \(f(x) = u(x) \cdot v(x)\), then \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\). 1. **Identify the two functions:** - \(u(x) = \sqrt{4 - 5x^3}\) - \(v(x) = \cot(8 - 4x)\) 2. **Differentiate \(u(x)\):** - Rewrite \(u(x)\) as \((4 - 5x^3)^{1/2}\). - Apply the chain rule: \(u'(x) = \frac{1}{2}(4 - 5x^3)^{-1/2} \cdot (-15x^2)\). - Simplify: \(u'(x) = -\frac{15x^2}{2\sqrt{4 - 5x^3}}\). 3. **Differentiate \(v(x)\):** - The derivative of \(\cot(z)\) is \(-\csc^2(z)\). - Apply the chain rule: \(v'(x) = -\csc^2(8 - 4x) \cdot (-4)\). - Simplify: \(v'(x) = 4\csc^2(8 - 4x)\). 4. **Apply the product rule:** - Compute \(u'(x) \cdot v(x) + u(x) \cdot v'(x)\): \[ f'(x) = \left( -\frac{15x^2}{2\sqrt{4 - 5x^3}} \right) \cot(8 - 4x) + \sqrt{4 - 5x^3} \cdot 4\csc^2(8 - 4x) \] 5. **Combine terms over a common denominator:** - The first term is already over \(\sqrt{4 - 5x^3}\). - Rewrite the second term over the same denominator: \[ 4\sqrt{4 - 5x^3} \csc^2(8 - 4x) = \frac{8(4 - 5x^3) \csc^2(8 - 4x)}{2\sqrt{4 - 5x^3}} \] - Combine both terms: \[ f'(x) = \frac{-15x^2 \cot(8 - 4x) + 8(4 - 5x^3)\csc^2(8 - 4x)}{2\sqrt{4 - 5x^3}} \] Thus, the final answer is: \[ \boxed{\dfrac{-15x^2 \cot(8 - 4x) + 8(4 - 5x^3)\csc^2(8 - 4x)}{2\sqrt{4 - 5x^3}}} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| gpto1 | To find the derivative of the function \( y = \sqrt{4 - 5x^3} \cot(8 - 4x) \), we'll use the product rule along with the chain rule. **Step 1:** Identify \( u(x) \) and \( v(x) \) Let \( u(x) = \sqrt{4 - 5x^3} = [4 - 5x^3]^{1/2} \) and \( v(x) = \cot(8 - 4x) \). **Step 2:** Find \( u'(x) \) Using the chain rule: \[ u'(x) = \dfrac{1}{2}[4 - 5x^3]^{-1/2} \cdot (-15x^2) = -\dfrac{15x^2}{2\sqrt{4 - 5x^3}} \] **Step 3:** Find \( v'(x) \) The derivative of \( \cot(u) \) is \( -\csc^2(u) \cdot u' \): \[ v'(x) = -\csc^2(8 - 4x) \cdot (-4) = 4\csc^2(8 - 4x) \] **Step 4:** Apply the product rule \[ y' = u'(x)v(x) + u(x)v'(x) \] Plugging in the expressions for \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ y' = \left( -\dfrac{15x^2}{2\sqrt{4 - 5x^3}} \right) \cot(8 - 4x) + \sqrt{4 - 5x^3} \cdot 4\csc^2(8 - 4x) \] **Final Answer:** \[ \boxed{\, y' = -\dfrac{15x^{2}}{2\sqrt{4-5x^{3}}}\cot(8-4x)\,+\,4\sqrt{4-5x^{3}}\,\csc^{2}(8-4x)\,} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| gpt-4o-2024-08-06 | To differentiate the function \( y = \sqrt{4 - 5x^3} \cot(8 - 4x) \), we will use the product rule and the chain rule. The product rule states that if \( y = u \cdot v \), then \( y' = u'v + uv' \). Let \( u = \sqrt{4 - 5x^3} \) and \( v = \cot(8 - 4x) \). First, we differentiate \( u = \sqrt{4 - 5x^3} \). \[ u = (4 - 5x^3)^{1/2} \] Using the chain rule, the derivative of \( u \) is: \[ u' = \frac{1}{2} (4 - 5x^3)^{-1/2} \cdot (-15x^2) = -\frac{15x^2}{2\sqrt{4 - 5x^3}} \] Next, we differentiate \( v = \cot(8 - 4x) \). The derivative of \( \cot(x) \) is \(-\csc^2(x)\), so using the chain rule: \[ v' = -\csc^2(8 - 4x) \cdot (-4) = 4 \csc^2(8 - 4x) \] Now, apply the product rule: \[ y' = u'v + uv' \] Substitute \( u' \), \( u \), \( v' \), and \( v \): \[ y' = \left(-\frac{15x^2}{2\sqrt{4 - 5x^3}}\right) \cot(8 - 4x) + \sqrt{4 - 5x^3} \cdot 4 \csc^2(8 - 4x) \] Simplify the expression: \[ y' = -\frac{15x^2 \cot(8 - 4x)}{2\sqrt{4 - 5x^3}} + 4\sqrt{4 - 5x^3} \csc^2(8 - 4x) \] Thus, the derivative of the function is: \[ \boxed{-\frac{15x^2 \cot(8 - 4x)}{2\sqrt{4 - 5x^3}} + 4\sqrt{4 - 5x^3} \csc^2(8 - 4x)} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| DeepSeek-Coder-V2-Lite-Instruct | To differentiate the function \(\sqrt{4-5 x^3} \cot (8-4 x)\), we will use the product rule. The product rule states that if we have a product of two functions \(u(x)\) and \(v(x)\), the derivative of their product is given by: \[ (uv)' = u'v + uv' \] Let \(u(x) = \sqrt{4-5 x^3}\) and \(v(x) = \cot (8-4 x)\). First, we need to find the derivatives of \(u(x)\) and \(v(x)\). ### Step 1: Differentiate \(u(x) = \sqrt{4-5 x^3}\) Using the chain rule, we have: \[ u(x) = (4-5 x^3)^{1/2} \] The derivative of \(u(x)\) is: \[ u'(x) = \frac{1}{2} (4-5 x^3)^{-1/2} \cdot (-15 x^2) = -\frac{15 x^2}{2 \sqrt{4-5 x^3}} \] ### Step 2: Differentiate \(v(x) = \cot (8-4 x)\) Using the chain rule, we have: \[ v(x) = \cot (8-4 x) \] The derivative of \(v(x)\) is: \[ v'(x) = -\csc^2 (8-4 x) \cdot (-4) = 4 \csc^2 (8-4 x) \] ### Step 3: Apply the product rule Now, we apply the product rule: \[ \frac{d}{dx} \left( \sqrt{4-5 x^3} \cot (8-4 x) \right) = u'(x) v(x) + u(x) v'(x) \] Substituting the derivatives we found: \[ \frac{d}{dx} \left( \sqrt{4-5 x^3} \cot (8-4 x) \right) = \left( -\frac{15 x^2}{2 \sqrt{4-5 x^3}} \right) \cot (8-4 x) + \sqrt{4-5 x^3} \left( 4 \csc^2 (8-4 x) \right) \] ### Step 4: Simplify the expression Combining the terms, we get: \[ \frac{d}{dx} \left( \sqrt{4-5 x^3} \cot (8-4 x) \right) = -\frac{15 x^2 \cot (8-4 x)}{2 \sqrt{4-5 x^3}} + 4 \sqrt{4-5 x^3} \csc^2 (8-4 x) \] Thus, the final answer is: \[ \boxed{-\frac{15 x^2 \cot (8-4 x)}{2 \sqrt{4-5 x^3}} + 4 \sqrt{4-5 x^3} \csc^2 (8-4 x)} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $\sqrt{4-5 x^3} \cot (8-4 x)$. This is a product of two functions, so we will need to use the product rule for differentiation. ## Step 2: Apply the product rule for differentiation The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. In this case, $u(x) = \sqrt{4-5 x^3}$ and $v(x) = \cot (8-4 x)$. ## Step 3: Differentiate the first function, $u(x) = \sqrt{4-5 x^3}$ To differentiate $\sqrt{4-5 x^3}$, we can use the chain rule. The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, so the derivative of $\sqrt{4-5 x^3}$ is $\frac{1}{2\sqrt{4-5 x^3}} \cdot (-15x^2)$. ## Step 4: Differentiate the second function, $v(x) = \cot (8-4 x)$ To differentiate $\cot (8-4 x)$, we can use the chain rule. The derivative of $\cot x$ is $-\csc^2 x$, so the derivative of $\cot (8-4 x)$ is $-\csc^2 (8-4 x) \cdot (-4)$. ## Step 5: Substitute the derivatives into the product rule formula Substituting $u'(x) = \frac{1}{2\sqrt{4-5 x^3}} \cdot (-15x^2)$ and $v'(x) = -\csc^2 (8-4 x) \cdot (-4)$ into the product rule formula, we get: $f'(x) = \left(\frac{1}{2\sqrt{4-5 x^3}} \cdot (-15x^2)\right)\cot (8-4 x) + \sqrt{4-5 x^3} \left(-\csc^2 (8-4 x) \cdot (-4)\right)$. ## Step 6: Simplify the derivative Simplifying the derivative, we get: $f'(x) = \frac{-15x^2}{2\sqrt{4-5 x^3}} \cot (8-4 x) + 4\sqrt{4-5 x^3} \csc^2 (8-4 x)$. The final answer is: $\boxed{\frac{-15x^2}{2\sqrt{4-5 x^3}} \cot (8-4 x) + 4\sqrt{4-5 x^3} \csc^2 (8-4 x)}$ |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $\sqrt{4-5 x^3} \cot (8-4 x)$. This is a product of two functions, so we will need to use the product rule of differentiation. ## Step 2: Apply the product rule of differentiation The product rule states that if we have a function of the form $u(x)v(x)$, then the derivative is given by $u'(x)v(x) + u(x)v'(x)$. In this case, $u(x) = \sqrt{4-5 x^3}$ and $v(x) = \cot (8-4 x)$. ## Step 3: Differentiate the first function, $u(x) = \sqrt{4-5 x^3}$ To differentiate $u(x)$, we will use the power rule and the sum rule. The derivative of $\sqrt{4-5 x^3}$ is $\frac{1}{2}(4-5 x^3)^{-\frac{1}{2}} \cdot (-15x^2)$. ## Step 4: Differentiate the second function, $v(x) = \cot (8-4 x)$ To differentiate $v(x)$, we will use the chain rule. The derivative of $\cot (8-4 x)$ is $-\csc^2 (8-4 x) \cdot (-4)$. ## Step 5: Apply the product rule Now we can apply the product rule by multiplying the derivative of $u(x)$ by $v(x)$ and adding the product of $u(x)$ and the derivative of $v(x)$. This gives us $\frac{1}{2}(4-5 x^3)^{-\frac{1}{2}} \cdot (-15x^2) \cot (8-4 x) + \sqrt{4-5 x^3} \cdot (-\csc^2 (8-4 x) \cdot (-4))$. ## Step 6: Simplify the expression Simplifying the expression, we get $\frac{-15x^2 \cot (8-4 x)}{2\sqrt{4-5 x^3}} + \frac{4\sqrt{4-5 x^3}}{\sin^2 (8-4 x)}$. ## Step 7: Write the final answer in the required format $\boxed{\frac{-15x^2 \cot (8-4 x)}{2\sqrt{4-5 x^3}} + \frac{4\sqrt{4-5 x^3}}{\sin^2 (8-4 x)}}$ The final answer is: $\boxed{\frac{-15x^2 \cot (8-4 x)}{2\sqrt{4-5 x^3}} + \frac{4\sqrt{4-5 x^3}}{\sin^2 (8-4 x)}}$ |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| Qwen2-72B-Instruct | To differentiate the given function, we will use the product rule and the chain rule. The function is: \[f(x) = \sqrt{4-5x^3}\cot(8-4x)\] Let's denote the two parts of the product as: \[u(x) = \sqrt{4-5x^3}\] \[v(x) = \cot(8-4x)\] First, we find the derivatives of \(u(x)\) and \(v(x)\): \[ u'(x) = \frac{d}{dx}\left(\sqrt{4-5x^3}\right) = \frac{1}{2}(4-5x^3)^{-\frac{1}{2}}(-15x^2) = -\frac{15x^2}{2\sqrt{4-5x^3}} \] \[ v'(x) = \frac{d}{dx}\left(\cot(8-4x)\right) = -\csc^2(8-4x)(-4) = 4\csc^2(8-4x) \] Now, using the product rule, which states that if \(f(x) = u(x)v(x)\), then \(f'(x) = u(x)v'(x) + u'(x)v(x)\), we can find the derivative of \(f(x)\): \[ f'(x) = u(x)v'(x) + u'(x)v(x) \] Substituting \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\) into the equation: \[ f'(x) = \sqrt{4-5x^3}\cdot4\csc^2(8-4x) - \frac{15x^2}{2\sqrt{4-5x^3}}\cdot\cot(8-4x) \] Simplifying this expression: \[ f'(x) = 4\sqrt{4-5x^3}\csc^2(8-4x) - \frac{15x^2\cot(8-4x)}{2\sqrt{4-5x^3}} \] This is the derivative of the given function. To present it in the requested format: \[ \boxed{f'(x) = 4\sqrt{4-5x^3}\csc^2(8-4x) - \frac{15x^2\cot(8-4x)}{2\sqrt{4-5x^3}}} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| baichuan4 | To differentiate the function $f(x) = \sqrt{4-5x^3} \cot(8-4x)$, we will use the product rule and the chain rule. The product rule states that if $u(x)$ and $v(x)$ are functions, then the derivative of their product is given by: $$ (uv)' = u'v + uv' $$ The chain rule states that if $g(x)$ is a function and $h(u)$ is a function of $u$, then the derivative of $h(g(x))$ is given by: $$ \frac{d}{dx}h(g(x)) = h'(g(x)) \cdot g'(x) $$ First, we will identify $u(x)$ and $v(x)$ in our function: $$ u(x) = \sqrt{4-5x^3} $$ $$ v(x) = \cot(8-4x) $$ Now, we will find the derivatives of $u(x)$ and $v(x)$. For $u(x)$, we apply the chain rule. Let $g(x) = 4-5x^3$ and $h(u) = \sqrt{u}$. Then: $$ u'(x) = h'(g(x)) \cdot g'(x) $$ $$ h'(u) = \frac{1}{2\sqrt{u}} $$ $$ g'(x) = -15x^2 $$ $$ u'(x) = \frac{1}{2\sqrt{4-5x^3}} \cdot (-15x^2) $$ $$ u'(x) = -\frac{15x^2}{2\sqrt{4-5x^3}} $$ For $v(x)$, we also apply the chain rule. Let $g(x) = 8-4x$ and $h(u) = \cot(u)$. Then: $$ v'(x) = h'(g(x)) \cdot g'(x) $$ $$ h'(u) = -\csc^2(u) $$ $$ g'(x) = -4 $$ $$ v'(x) = -\csc^2(8-4x) \cdot (-4) $$ $$ v'(x) = 4\csc^2(8-4x) $$ Now we can apply the product rule to find the derivative of $f(x)$: $$ f'(x) = u'(x)v(x) + u(x)v'(x) $$ $$ f'(x) = \left(-\frac{15x^2}{2\sqrt{4-5x^3}}\right)\cot(8-4x) + \sqrt{4-5x^3}(4\csc^2(8-4x)) $$ Simplifying, we get: $$ f'(x) = -\frac{15x^2}{2\sqrt{4-5x^3}}\cot(8-4x) + \frac{4(4-5x^3)}{\sqrt{4-5x^3}}\csc^2(8-4x) $$ We can further simplify by combining the terms: $$ f'(x) = \frac{-15x^2\cot(8-4x) + 8(4-5x^3)\csc^2(8-4x)}{2\sqrt{4-5x^3}} $$ Finally, we can write the derivative in a box: $$ \boxed{f'(x) = \frac{-15x^2\cot(8-4x) + 8(4-5x^3)\csc^2(8-4x)}{2\sqrt{4-5x^3}}} $$ |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| doubao | Let $y=\sqrt{4-5x^3}\cot(8-4x)$. Using the product rule and the chain rule: \[ \begin{align*} y'&=(\sqrt{4-5x^3})'\cot(8-4x)+\sqrt{4-5x^3}(\cot(8-4x))'\\ &=\frac{1}{2}(4-5x^3)^{-\frac{1}{2}}\times(-15x^2)\cot(8-4x)-\sqrt{4-5x^3}\csc^2(8-4x)\times(-4)\\ &=\frac{-15x^2}{2\sqrt{4-5x^3}}\cot(8-4x)+4\sqrt{4-5x^3}\csc^2(8-4x)\\ \end{align*} \] So the answer is $\boxed{\frac{-15x^2}{2\sqrt{4-5x^3}}\cot(8-4x)+4\sqrt{4-5x^3}\csc^2(8-4x)}$. |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| gpto1 | To find the derivative of the function \( y = \sqrt{4 - 5x^3} \cot(8 - 4x) \), we'll use the product rule along with the chain rule. **Step 1:** Identify \( u(x) \) and \( v(x) \) Let \( u(x) = \sqrt{4 - 5x^3} = [4 - 5x^3]^{1/2} \) and \( v(x) = \cot(8 - 4x) \). **Step 2:** Find \( u'(x) \) Using the chain rule: \[ u'(x) = \dfrac{1}{2}[4 - 5x^3]^{-1/2} \cdot (-15x^2) = -\dfrac{15x^2}{2\sqrt{4 - 5x^3}} \] **Step 3:** Find \( v'(x) \) The derivative of \( \cot(u) \) is \( -\csc^2(u) \cdot u' \): \[ v'(x) = -\csc^2(8 - 4x) \cdot (-4) = 4\csc^2(8 - 4x) \] **Step 4:** Apply the product rule \[ y' = u'(x)v(x) + u(x)v'(x) \] Plugging in the expressions for \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ y' = \left( -\dfrac{15x^2}{2\sqrt{4 - 5x^3}} \right) \cot(8 - 4x) + \sqrt{4 - 5x^3} \cdot 4\csc^2(8 - 4x) \] **Final Answer:** \[ \boxed{\, y' = -\dfrac{15x^{2}}{2\sqrt{4-5x^{3}}}\cot(8-4x)\,+\,4\sqrt{4-5x^{3}}\,\csc^{2}(8-4x)\,} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| kimi | To differentiate the function \( f(x) = \sqrt{4-5x^3} \cot(8-4x) \), we will use the product rule and the chain rule. The product rule states that if \( f(x) = g(x)h(x) \), then \( f'(x) = g'(x)h(x) + g(x)h'(x) \). Here, let \( g(x) = \sqrt{4-5x^3} \) and \( h(x) = \cot(8-4x) \). First, we find the derivatives of \( g(x) \) and \( h(x) \). 1. Differentiate \( g(x) = \sqrt{4-5x^3} \): \[ g(x) = (4-5x^3)^{1/2} \] Using the chain rule, let \( u = 4-5x^3 \). Then \( g(x) = u^{1/2} \) and \( g'(x) = \frac{1}{2}u^{-1/2} \cdot u' \). \[ u' = -15x^2 \] Therefore, \[ g'(x) = \frac{1}{2}(4-5x^3)^{-1/2} \cdot (-15x^2) = -\frac{15x^2}{2\sqrt{4-5x^3}} \] 2. Differentiate \( h(x) = \cot(8-4x) \): \[ h'(x) = -\csc^2(8-4x) \cdot (-4) = 4\csc^2(8-4x) \] Now, apply the product rule: \[ f'(x) = g'(x)h(x) + g(x)h'(x) \] Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \): \[ f'(x) = \left( -\frac{15x^2}{2\sqrt{4-5x^3}} \right) \cot(8-4x) + \sqrt{4-5x^3} \cdot 4\csc^2(8-4x) \] Simplify the expression: \[ f'(x) = -\frac{15x^2 \cot(8-4x)}{2\sqrt{4-5x^3}} + 4\sqrt{4-5x^3} \csc^2(8-4x) \] Thus, the final answer is: \[ \boxed{-\frac{15x^2 \cot(8-4x)}{2\sqrt{4-5x^3}} + 4\sqrt{4-5x^3} \csc^2(8-4x)} \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| minimax | To differentiate the function $f(x) = \sqrt{4-5x^3} \cot (8-4x)$, we will use the product rule and the chain rule. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. The chain rule is used to differentiate composite functions. Let's break down the function into two parts: $u(x) = \sqrt{4-5x^3}$ and $v(x) = \cot (8-4x)$. First, we find the derivatives of $u(x)$ and $v(x)$ separately. For $u(x)$, we have $u(x) = (4-5x^3)^{1/2}$. Using the chain rule, the derivative $u'(x)$ is: $$ u'(x) = \frac{1}{2}(4-5x^3)^{-1/2} \cdot (-15x^2) = -\frac{15x^2}{2\sqrt{4-5x^3}} $$ For $v(x)$, we have $v(x) = \ |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| wenxinyiyan | The given function is $f(x) = \sqrt{4-5 x^3} \cot (8-4 x)$. First, we need to find the derivatives of $\sqrt{4-5 x^3}$ and $\cot(8-4 x)$ separately. For $\sqrt{4-5 x^3}$, let's denote it as $u(x) = \sqrt{4-5 x^3}$. Using the chain rule, we have $$u'(x) = \frac{d}{dx}\sqrt{4-5 x^3} = \frac{-15x^2}{2\sqrt{4-5x^3}}.$$ For $\cot(8-4 x)$, let's denote it as $v(x) = \cot(8-4 x)$. The derivative of $\cot(x)$ is $-\csc^2(x)$, so using the chain rule again, we get $$v'(x) = \frac{d}{dx}\cot(8-4 x) = 4\csc^2(8-4x).$$ Now, we use the product rule to find the derivative of $f(x)$: $$f'(x) = u'(x)v(x) + u(x)v'(x).$$ Substituting $u'(x)$ and $v'(x)$ into the product rule, we obtain: $$f'(x) = \frac{-15x^2}{2\sqrt{4-5x^3}} \cdot \cot(8-4 x) + \sqrt{4-5 x^3} \cdot 4\csc^2(8-4x).$$ Simplifying, we get: $$f'(x) = \frac{-15x^2\cot(8-4x)}{2\sqrt{4-5x^3}} + 4\sqrt{4-5x^3}\csc^2(8-4x).$$ Therefore, the final answer is $$\boxed{f'(x) = \frac{-15x^2\cot(8-4x)}{2\sqrt{4-5x^3}} + 4\sqrt{4-5x^3}\csc^2(8-4x)}.$$ |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| zhipu | To differentiate the function \( f(x) = \sqrt{4 - 5x^3} \cot(8 - 4x) \), we will use the product rule and the chain rule. The product rule states that if \( u(x) \) and \( v(x) \) are differentiable functions, then the derivative of their product \( u(x)v(x) \) is given by: \[ (uv)' = u'v + uv'. \] Let \( u(x) = \sqrt{4 - 5x^3} \) and \( v(x) = \cot(8 - 4x) \). First, we find the derivatives of \( u(x) \) and \( v(x) \) separately. ### Derivative of \( u(x) = \sqrt{4 - 5x^3} \) Using the chain rule, let \( g(x) = 4 - 5x^3 \) and \( h(x) = \sqrt{x} \). Then \( u(x) = h(g(x)) \). The derivative of \( h(x) = \sqrt{x} \) is: \[ h'(x) = \frac{1}{2\sqrt{x}}. \] The derivative of \( g(x) = 4 - 5x^3 \) is: \[ g'(x) = -15x^2. \] Applying the chain rule: \[ u'(x) = h'(g(x)) \cdot g'(x) = \frac{1}{2\sqrt{4 - 5x^3}} \cdot (-15x^2) = \frac{-15x^2}{2\sqrt{4 - 5x^3}}. \] ### Derivative of \( v(x) = \cot(8 - 4x) \) Using the chain rule again, let \( p(x) = 8 - 4x \) and \( q(x) = \cot(x) \). Then \( v(x) = q(p(x)) \). The derivative of \( q(x) = \cot(x) \) is: \[ q'(x) = -\csc^2(x). \] The derivative of \( p(x) = 8 - 4x \) is: \[ p'(x) = -4. \] Applying the chain rule: \[ v'(x) = q'(p(x)) \cdot p'(x) = -\csc^2(8 - 4x) \cdot (-4) = 4\csc^2(8 - 4x). \] ### Applying the Product Rule Now, using the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x). \] Substituting \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ f'(x) = \left( \frac{-15x^2}{2\sqrt{4 - 5x^3}} \right) \cot(8 - 4x) + \sqrt{4 - 5x^3} \left( 4\csc^2(8 - 4x) \right). \] Simplifying the expression: \[ f'(x) = \frac{-15x^2 \cot(8 - 4x)}{2\sqrt{4 - 5x^3}} + 4\sqrt{4 - 5x^3} \csc^2(8 - 4x). \] Thus, the derivative of the function \( \sqrt{4 - 5x^3} \cot(8 - 4x) \) is: \[ \boxed{\frac{-15x^2 \cot(8 - 4x)}{2\sqrt{4 - 5x^3}} + 4\sqrt{4 - 5x^3} \csc^2(8 - 4x)}. \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |
| deepseek-v3 | To differentiate the function \( \sqrt{4 - 5x^3} \cdot \cot(8 - 4x) \), we'll use the **Product Rule** and the **Chain Rule**. Here's a step-by-step solution: 1. **Identify the Components:** \[ f(x) = \sqrt{4 - 5x^3} = (4 - 5x^3)^{1/2} \] \[ g(x) = \cot(8 - 4x) \] 2. **Differentiate \( f(x) \) using the Chain Rule:** \[ f'(x) = \frac{1}{2}(4 - 5x^3)^{-1/2} \cdot (-15x^2) = \frac{-15x^2}{2\sqrt{4 - 5x^3}} \] 3. **Differentiate \( g(x) \) using the Chain Rule:** \[ g'(x) = -\csc^2(8 - 4x) \cdot (-4) = 4\csc^2(8 - 4x) \] 4. **Apply the Product Rule:** \[ \frac{d}{dx} [f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) \] Substituting the derivatives: \[ \frac{d}{dx} = \left( \frac{-15x^2}{2\sqrt{4 - 5x^3}} \right) \cdot \cot(8 - 4x) + \sqrt{4 - 5x^3} \cdot 4\csc^2(8 - 4x) \] 5. **Simplify the Expression:** \[ \frac{d}{dx} = -\frac{15x^2 \cot(8 - 4x)}{2\sqrt{4 - 5x^3}} + 4\sqrt{4 - 5x^3} \cdot \csc^2(8 - 4x) \] **Final Answer:** \[ \boxed{ -\frac{15x^2 \cot(8 - 4x)}{2\sqrt{4 - 5x^3}} + 4\sqrt{4 - 5x^3} \cdot \csc^2(8 - 4x) } \] |
-\frac{\left(80 x^3+15 x^2 \sin (16-8 x)-64\right) \csc ^2(8-4 x)}{4 \sqrt{4-5 x^3}} |