| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the derivative of \( z \) with respect to \( x \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function \( z(x) = \frac{f(x)}{g(x)} \), then its derivative \( z'(x) \) is given by: \[ z'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] In this problem, \( f(x) = 1 + e^{-2x} \) and \( g(x) = x + \tan(12x) \). We need to find \( f'(x) \) and \( g'(x) \). 1. Find \( f'(x) \): \[ f(x) = 1 + e^{-2x} \] Differentiating \( f(x) \) with respect to \( x \): \[ f'(x) = -2e^{-2x} \] 2. Find \( g'(x) \): \[ g(x) = x + \tan(12x) \] Differentiating \( g(x) \) with respect to \( x \): \[ g'(x) = 1 + 12\sec^2(12x) \] Now, we can use the quotient rule to find \( \frac{\partial z}{\partial x} \): \[ \frac{\partial z}{\partial x} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] Substituting \( f(x) \), \( f'(x) \), \( g(x) \), and \( g'(x) \): \[ \frac{\partial z}{\partial x} = \frac{(-2e^{-2x})(x + \tan(12x)) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} \] We want to find the derivative at \( x = 1 \): \[ \left.\frac{\partial z}{\partial x}\right|_{x=1} = \frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2} \] This expression can be evaluated using a calculator or software, as \( \tan(12) \) and \( \sec(12) \) cannot be simplified without numerical methods. Note that the values of \( \tan(12) \) and \( \sec(12) \) are not real numbers, as \( 12 \) radians is outside of the principal value range for these functions, and they will result in complex numbers. However, the expression for the derivative \( \frac{\partial z}{\partial x} \) is valid and can be evaluated numerically. |
-153.59 |
| deepseekr1 | To find the derivative of \( z = \frac{1 + e^{-2x}}{x + \tan(12x)} \) at \( x = 1 \), we use the quotient rule. Given: \[ f(x) = 1 + e^{-2x}, \quad g(x) = x + \tan(12x) \] First, compute the derivatives: \[ f'(x) = -2e^{-2x}, \quad g'(x) = 1 + 12\sec^2(12x) \] Applying the quotient rule: \[ \frac{\partial z}{\partial x} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \] Substitute \( x = 1 \): \[ f'(1) = -2e^{-2}, \quad g(1) = 1 + \tan(12), \quad f(1) = 1 + e^{-2}, \quad g'(1) = 1 + 12\sec^2(12) \] Compute each term numerically: - \( e^{-2} \approx 0.1353 \) - \( \tan(12) \approx -1.5727 \) - \( \sec(12) \approx 1.1853 \), so \( \sec^2(12) \approx 1.404 \) Calculate the numerator: \[ (-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12)) \approx 0.1550 - 20.2775 = -20.1225 \] Denominator: \[ (1 + \tan(12))^2 \approx (-0.5727)^2 \approx 0.3280 \] Thus, the derivative at \( x = 1 \): \[ \frac{-20.1225}{0.3280} \approx -61.36 \] **Answer:** \(\boxed{-61.36}\) |
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| gpto1 | To find the derivative \(\frac{\partial z}{\partial x}\) of the function \[ z = \frac{1 + e^{-2x}}{x + \tan(12x)} \] at \( x = 1 \), we can follow these steps: **Step 1: Identify and simplify the numerator and denominator functions.** Let’s denote: - \( N(x) = 1 + e^{-2x} \) (the numerator), - \( D(x) = x + \tan(12x) \) (the denominator). Then: \[ z = \frac{N(x)}{D(x)} \] **Step 2: Compute the derivatives of the numerator and denominator.** Compute \( N'(x) \) and \( D'(x) \): 1. \( N'(x) = \frac{d}{dx}[1 + e^{-2x}] = -2e^{-2x} \). 2. \( D'(x) = \frac{d}{dx}[x + \tan(12x)] = 1 + 12\sec^2(12x) \). **Step 3: Use the quotient rule to find \( \frac{dz}{dx} \).** The quotient rule states: \[ \frac{d}{dx}\left(\frac{N(x)}{D(x)}\right) = \frac{N'(x)D(x) - N(x)D'(x)}{[D(x)]^2} \] So: \[ \frac{dz}{dx} = \frac{[-2e^{-2x}][x + \tan(12x)] - [1 + e^{-2x}][1 + 12\sec^2(12x)]}{[x + \tan(12x)]^2} \] **Step 4: Evaluate the numerator and denominator at \( x = 1 \).** We need to compute the values of \( N(1) \), \( N'(1) \), \( D(1) \), and \( D'(1) \). 1. Compute \( e^{-2(1)} = e^{-2} \approx 0.1353 \). 2. Then: - \( N(1) = 1 + e^{-2} \approx 1 + 0.1353 = 1.1353 \). - \( N'(1) = -2e^{-2} \approx -2 \times 0.1353 = -0.2707 \). 3. Compute \( \tan(12(1)) = \tan(12) \) and \( \sec^2(12(1)) = \sec^2(12) \). - Since 12 radians is not a standard angle, we'll compute these numerically. 4. Adjusting the angle: - \( 12 \) radians is equivalent to \( 12 \mod \pi \) because the tangent function has a period of \( \pi \). - \( \frac{12}{\pi} \approx 3.8197 \), so \( 12 - 3\pi \approx 0.8197\pi \). - Therefore, \( \tan(12) = \tan(0.8197\pi) \). 5. Compute \( \tan(0.8197\pi) \): - \( 0.8197\pi \) corresponds to \( 147.55^\circ \). - \( \tan(147.55^\circ) = -\tan(32.45^\circ) \approx -0.6357 \). 6. Compute \( \sec^2(12) \): - \( \cos(147.55^\circ) \approx -0.8401 \). - \( \sec(12) = \frac{1}{\cos(12)} = \frac{1}{-0.8401} = -1.1904 \). - \( \sec^2(12) = [-1.1904]^2 = 1.4171 \). 7. Then: - \( D(1) = 1 + \tan(12) \approx 1 + (-0.6357) = 0.3643 \). - \( D'(1) = 1 + 12\sec^2(12) \approx 1 + 12 \times 1.4171 = 1 + 17.0052 = 18.0052 \). **Step 5: Compute the numerator and denominator of the derivative at \( x = 1 \).** Compute: \[ \text{Numerator} = N'(1)D(1) - N(1)D'(1) = (-0.2707)(0.3643) - (1.1353)(18.0052) \] \[ \text{Numerator} \approx -0.0986 - 20.4325 = -20.5311 \] \[ \text{Denominator} = [D(1)]^2 = [0.3643]^2 = 0.1327 \] **Step 6: Compute \( \frac{dz}{dx} \) at \( x = 1 \).** \[ \frac{dz}{dx} \Big|_{x=1} = \frac{\text{Numerator}}{\text{Denominator}} = \frac{-20.5311}{0.1327} \] \[ \frac{dz}{dx} \Big|_{x=1} \approx -154.7783 \] **Answer:** Approximately \(-154.78\); that is, \(\dfrac{\partial z}{\partial x}\bigg|_{x=1} \approx -154.78\) |
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| GPT-4o | To find the derivative \(\frac{\partial z}{\partial x}\) at \(x = 1\) for the function \(z = \frac{1 + e^{-2x}}{x + \tan(12x)}\), we will use the quotient rule and then evaluate it at \(x = 1\). The quotient rule states that if you have a function \(z = \frac{f(x)}{g(x)}\), then the derivative is given by: \[ \frac{dz}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \] Here, \(f(x) = 1 + e^{-2x}\) and \(g(x) = x + \tan(12x)\). First, we need to find \(f'(x)\) and \(g'(x)\). ### Step 1: Differentiate \(f(x)\) \[ f(x) = 1 + e^{-2x} \] \[ f'(x) = \frac{d}{dx}(1 + e^{-2x}) = 0 + \frac{d}{dx}(e^{-2x}) = -2e^{-2x} \] ### Step 2: Differentiate \(g(x)\) \[ g(x) = x + \tan(12x) \] \[ g'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\tan(12x)) = 1 + 12\sec^2(12x) \] ### Step 3: Apply the Quotient Rule \[ z = \frac{1 + e^{-2x}}{x + \tan(12x)} \] \[ \frac{dz}{dx} = \frac{(1 + \tan(12x))(-2e^{-2x}) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} \] ### Step 4: Evaluate at \(x = 1\) First, compute the values of \(f(1)\), \(f'(1)\), \(g(1)\), and \(g'(1)\): \[ f(1) = 1 + e^{-2 \cdot 1} = 1 + e^{-2} \] \[ f'(1) = -2e^{-2} \] \[ g(1) = 1 + \tan(12 \cdot 1) = 1 + \tan(12) \] \[ g'(1) = 1 + 12\sec^2(12) \] Now substitute these into the quotient rule formula: \[ \frac{dz}{dx}\Bigg|_{x=1} = \frac{(1 + \tan(12))(-2e^{-2}) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2} \] Simplify the numerator: \[ \text{Numerator} = -2e^{-2}(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12)) \] \[ = -2e^{-2} - 2e^{-2}\tan(12) - 1 - e^{-2} - 12\sec^2(12) - 12e^{-2}\sec^2(12) \] \[ = -3e^{-2} - 2e^{-2}\tan(12) - 1 - 12\sec^2(12) - 12e^{-2}\sec^2(12) \] So the final expression for \(\frac{dz}{dx}\) at \(x = 1\) is: \[ \frac{dz}{dx}\Bigg|_{x=1} = \frac{-3e^{-2} - 2e^{-2}\tan(12) - 1 - 12\sec^2(12) - 12e^{-2}\sec^2(12)}{(1 + \tan(12))^2} \] This expression gives the derivative of \(z\) with respect to \(x\) evaluated at \(x = 1\). |
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| llama3.1-70B-Instruct | To find the derivative of z with respect to x, we'll apply the quotient rule of differentiation, which states that if we have a function of the form: f(x) = \frac{g(x)}{h(x)} Then the derivative of f(x) is given by: f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{h(x)^2} In this case, we have: z = \frac{1 + e^{-2x}}{x + tan(12x)} So, we'll apply the quotient rule to find the derivative of z with respect to x. First, let's find the derivatives of the numerator and denominator: g(x) = 1 + e^{-2x} --> g'(x) = -2e^{-2x} h(x) = x + tan(12x) --> h'(x) = 1 + 12sec^2(12x) Now, we can plug these derivatives into the quotient rule formula: \frac{\partial z}{\partial x} = \frac{(x + tan(12x))(-2e^{-2x}) - (1 + e^{-2x})(1 + 12sec^2(12x))}{(x + tan(12x))^2} To find the value of the derivative at x = 1, we'll plug in x = 1 into the expression: \frac{\partial z}{\partial x} (1) = \frac{(1 + tan(12))( -2e^{-2}) - (1 + e^{-2})(1 + 12sec^2(12))}{(1 + tan(12))^2} Using a calculator to evaluate the trigonometric functions and exponentials, we get: \frac{\partial z}{\partial x} (1) ≈ -0.0677 So, the derivative of z with respect to x at x = 1 is approximately -0.0677. |
-153.59 |
| llama3.1-405B-Instruct | To find the derivative of z with respect to x, we can use the quotient rule and the chain rule. ## Step 1: Define the function z z = \frac{1 + e^{-2x}}{x + tan(12x)} ## Step 2: Apply the quotient rule to find the derivative of z The quotient rule states that if z = \frac{f(x)}{g(x)}, then \frac{\partial z}{\partial x} = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}. ## Step 3: Identify f(x) and g(x) f(x) = 1 + e^{-2x} g(x) = x + tan(12x) ## Step 4: Find the derivatives of f(x) and g(x) f'(x) = -2e^{-2x} g'(x) = 1 + 12sec^2(12x) ## Step 5: Substitute f(x), g(x), f'(x), and g'(x) into the quotient rule formula \frac{\partial z}{\partial x} = \frac{(x + tan(12x))(-2e^{-2x}) - (1 + e^{-2x})(1 + 12sec^2(12x))}{(x + tan(12x))^2} ## Step 6: Simplify the expression for the derivative of z \frac{\partial z}{\partial x} = \frac{-2xe^{-2x} - 2e^{-2x}tan(12x) - 1 - 12sec^2(12x) - e^{-2x} - 12e^{-2x}sec^2(12x)}{(x + tan(12x))^2} ## Step 7: Evaluate the derivative of z at x = 1 \frac{\partial z}{\partial x} at x = 1 is \frac{-2e^{-2} - 2e^{-2}tan(12) - 1 - 12sec^2(12) - e^{-2} - 12e^{-2}sec^2(12)}{(1 + tan(12))^2} ## Step 8: Calculate the numerical value of the derivative of z at x = 1 Using a calculator, we get: -2e^{-2} - 2e^{-2}tan(12) - 1 - 12sec^2(12) - e^{-2} - 12e^{-2}sec^2(12) ≈ -3.718 (1 + tan(12))^2 ≈ 1.043 ## Step 9: Calculate the final value of the derivative of z at x = 1 \frac{\partial z}{\partial x} at x = 1 ≈ \frac{-3.718}{1.043} ≈ -3.565 The final answer is: $\boxed{-3.565}$ |
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| DeepSeek-Coder-V2 | To solve the problem of finding the derivative of \( z = \frac{1 + e^{-2x}}{x + \tan(12x)} \) with respect to \( x \) at \( x = 1 \), we will use the quotient rule for differentiation. The quotient rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of \( \frac{u(x)}{v(x)} \) is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] Here, let \( u(x) = 1 + e^{-2x} \) and \( v(x) = x + \tan(12x) \). We need to find the derivatives \( u'(x) \) and \( v'(x) \). First, compute \( u'(x) \): \[ u(x) = 1 + e^{-2x} \] \[ u'(x) = \frac{d}{dx}(1 + e^{-2x}) = 0 + (-2e^{-2x}) = -2e^{-2x} \] Next, compute \( v'(x) \): \[ v(x) = x + \tan(12x) \] \[ v'(x) = \frac{d}{dx}(x + \tan(12x)) = 1 + 12\sec^2(12x) \] Now, apply the quotient rule: \[ \frac{\partial z}{\partial x} = \frac{u'v - uv'}{v^2} \] Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule: \[ \frac{\partial z}{\partial x} = \frac{(-2e^{-2x})(x + \tan(12x)) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} \] Evaluate this expression at \( x = 1 \): \[ u(1) = 1 + e^{-2(1)} = 1 + e^{-2} \] \[ u'(1) = -2e^{-2(1)} = -2e^{-2} \] \[ v(1) = 1 + \tan(12(1)) = 1 + \tan(12) \] \[ v'(1) = 1 + 12\sec^2(12) \] Substitute these values into the derivative expression: \[ \frac{\partial z}{\partial x} \bigg|_{x=1} = \frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2} \] Simplify the numerator: \[ (-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12)) \] This expression is quite complex, but it is the correct form for the derivative at \( x = 1 \). The final answer is: \[ \boxed{\frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2}} \] |
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| Qwen2-72B-Instruct | To find the derivative of $z$ with respect to $x$, we will use the quotient rule for differentiation. The quotient rule states that if you have a function $z(x) = \frac{f(x)}{g(x)}$, then its derivative $z'(x)$ is given by: \[ z'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] In this problem, $f(x) = 1 + e^{-2x}$ and $g(x) = x + \tan(12x)$. We need to find $f'(x)$ and $g'(x)$. 1. Find $f'(x)$: \[ f(x) = 1 + e^{-2x} \] Differentiating $f(x)$ with respect to $x$: \[ f'(x) = -2e^{-2x} \] 2. Find $g'(x)$: \[ g(x) = x + \tan(12x) \] Differentiating $g(x)$ with respect to $x$: \[ g'(x) = 1 + 12\sec^2(12x) \] Now, we can use the quotient rule to find $\frac{\partial z}{\partial x}$: \[ \frac{\partial z}{\partial x} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] Substituting $f(x)$, $f'(x)$, $g(x)$, and $g'(x)$: \[ \frac{\partial z}{\partial x} = \frac{(-2e^{-2x})(x + \tan(12x)) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} \] We want to find the derivative at $x = 1$: \[ \left.\frac{\partial z}{\partial x}\right|_{x=1} = \frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2} \] This expression can be evaluated using a calculator or software, as $\tan(12)$ and $\sec(12)$ cannot be simplified without numerical methods. Note that the values of $\tan(12)$ and $\sec(12)$ are not real numbers, as $12$ radians is outside of the principal value range for these functions, and they will result in complex numbers. However, the expression for the derivative $\frac{\partial z}{\partial x}$ is valid and can be evaluated numerically. |
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| baichuan4 | To find the derivative $\frac{\partial z}{\partial x}$ of $z = \frac{1 + e^{-2x}}{x + \tan(12x)}$ at $x = 1$, we will use the quotient rule for derivatives, which states that for a function $f(x)$ divided by $g(x)$, the derivative is given by: $$ \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} $$ In our case, $f(x) = 1 + e^{-2x}$ and $g(x) = x + \tan(12x)$. We need to find $f'(x)$ and $g'(x)$. First, let's differentiate $f(x)$: $$ f'(x) = \frac{d}{dx}(1 + e^{-2x}) = 0 + (-2)e^{-2x} = -2e^{-2x} $$ Now, let's differentiate $g(x)$. Note that we need to use the chain rule for the $\tan(12x)$ term: $$ g'(x) = \frac{d}{dx}(x + \tan(12x)) = 1 + \sec^2(12x) \cdot 12 = 1 + 12\sec^2(12x) $$ Now we can apply the quotient rule: $$ \frac{\partial z}{\partial x} = \frac{-2e^{-2x}(x + \tan(12x)) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} $$ Simplifying this expression: $$ \frac{\partial z}{\partial x} = \frac{-2xe^{-2x} - 2e^{-2x}\tan(12x) - 1 - e^{-2x} - 12 - 12e^{-2x}\sec^2(12x)}{(x + \tan(12x))^2} $$ $$ = \frac{-2xe^{-2x} - 2e^{-2x}\tan(12x) - 13 - e^{-2x}(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} $$ Now, we need to evaluate this derivative at $x = 1$: $$ \frac{\partial z}{\partial x}\Big|_{x=1} = \frac{-2e^{-2} - 2e^{-2}\tan(12) - 13 - e^{-2}(1 + 12\sec^2(12))}{(1 + \tan(12))^2} $$ This is the final answer, where you would plug in the values for $e^{-2}$, $\tan(12)$, and $\sec^2(12)$ to get the numerical result. |
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| doubao | First find the derivative of z with respect to x using the quotient rule and the chain rule: \[ \begin{align*} \frac{\partial z}{\partial x}&=\frac{(x+\tan(12x))(-2e^{-2x})-(1+e^{-2x})(1+12\sec^2(12x))}{(x+\tan(12x))^2} \end{align*} \] Then evaluate this derivative at x = 1: \[ \begin{align*} \frac{\partial z}{\partial x}\mid_{x=1}&=\frac{(1+\tan(12))(-2e^{-2})-(1+e^{-2})(1+12\sec^2(12))}{(1+\tan(12))^2} \end{align*} \] This is a very complicated expression and may require further simplification depending on the specific requirements. |
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| minimax | To find the derivative of $z$ with respect to $x$, we will use the quotient rule. The quotient rule states that if you have a function $z = \frac{f(x)}{g(x)}$, then the derivative $\frac{\partial z}{\partial x}$ is given by: $$ \frac{\partial z}{\partial x} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} $$ where $f(x) = 1 + e^{-2x}$ and $g(x) = x + \tan(12x)$. First, we need to find the derivatives of $f(x)$ and $g(x)$: For $f(x) = 1 + e^{-2x}$: $$ f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(e^{-2x}) = 0 - 2e^{-2x} = -2e^{-2x} $$ For $g(x) = x + \tan(12x)$: $$ g'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\tan(12x)) = 1 + 12\sec^2(12x) $$ Now, we apply the quotient rule: $$ \frac{\partial z}{\partial x} = \frac{(-2e^{-2x})(x + \tan(12x)) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} $$ Simplify the numerator: $$ \frac{\partial z}{\partial x} = \frac{-2xe^{-2x} - 2e^{-2x}\tan(12x) - 1 - e^{-2x} - 12e^{-2x}\sec^2(12x)}{(x + \tan(12x))^2} $$ Combine like terms: $$ \frac{\partial z}{\partial x} = \frac{-2xe^{-2x} - 2e^{-2x}\tan(12x) - 12e^{-2x}\sec^2(12x) - 1 - e^{-2x}}{(x + \tan(12x))^2} $$ $$ \frac{\partial z}{\partial x} = \frac{-2xe^{-2x} - 3e^{-2x} - 2e^{-2x}\tan(12x) - 12e^{-2x}\sec^2(12x)}{(x + \tan(12x))^2} $$ Now we need to evaluate this derivative at $x = 1$: $$ \frac{\partial z}{\partial x}\bigg|_{x=1} = \frac{-2(1)e^{-2(1)} - 3e^{-2(1)} - 2e^{-2(1)}\tan(12(1)) - 12e^{-2(1)}\sec^2(12(1))}{(1 + \tan(12(1)))^2} $$ $$ \frac{\partial z}{\partial x}\bigg|_{x=1} = \frac{-2e^{-2} - 3e^{-2} - 2e^{-2}\tan(12) - 12e^{-2}\sec^2(12)}{(1 + \tan(12))^2} $$ Now we can plug in the values for $e$, $\tan(12)$, and $\sec^2(12)$ to get the numerical value: $$ \frac{\partial z}{\partial x}\bigg|_{x=1} = \frac{-2e^{-2} - 3e^{-2} - 2e^{-2}\tan(12) - 12e^{-2}\sec^2(12)}{(1 + \tan(12))^2} $$ $$ \frac{\partial z}{\partial x}\bigg|_{x=1} = \frac{-5e^{-2} - 2e^{-2}\tan(12) - 12e^{-2}\sec^2(12)}{(1 + \tan(12))^2} $$ You can now use a calculator to find the numerical value of this expression. |
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| kimi | To find the derivative of \( z \) with respect to \( x \), we can use the quotient rule. The quotient rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) is given by: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] In this case, \( u(x) = 1 + e^{-2x} \) and \( v(x) = x + \tan(12x) \). Let's find the derivatives of \( u(x) \) and \( v(x) \): \[ u'(x) = \frac{d}{dx}(1 + e^{-2x}) = -2e^{-2x} \] \[ v'(x) = \frac{d}{dx}(x + \tan(12x)) = 1 + \frac{d}{dx}(\tan(12x)) \] To find the derivative of \( \tan(12x) \), we use the chain rule: \[ \frac{d}{dx}(\tan(12x)) = \sec^2(12x) \cdot \frac{d}{dx}(12x) = \sec^2(12x) \cdot 12 \] Now we have all the components to apply the quotient rule: \[ \frac{\partial z}{\partial x} = \frac{(-2e^{-2x})(x + \tan(12x)) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2} \] Now we substitute \( x = 1 \) into the derivative: \[ \frac{\partial z}{\partial x}(1) = \frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2} \] Now we can calculate the numerical values: \[ \frac{\partial z}{\partial x}(1) = \frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2} \] Let's calculate the numerical values for \( \tan(12) \), \( e^{-2} \), and \( \sec^2(12) \): \[ \tan(12) \approx \tan(12 - 1 \cdot 2\pi) \approx \tan(5.716815) \approx -0.635859 \] \[ e^{-2} \approx 0.135335 \] \[ \sec^2(12) = 1 + \tan^2(12) \approx 1 + (-0.635859)^2 \approx 1.403502 \] Now we can substitute these values into the expression: \[ \frac{\partial z}{\partial x}(1) = \frac{(-2 \cdot 0.135335)(1 - 0.635859) - (1 + 0.135335)(1 + 12 \cdot 1.403502)}{(1 - 0.635859)^2} \] \[ \frac{\partial z}{\partial x}(1) = \frac{(-0.27067)(0.364141) - (1.135335)(16.840024)}{(0.364141)^2} \] \[ \frac{\partial z}{\partial x}(1) = \frac{-0.098535 - 19.141963}{0.132665} \] \[ \frac{\partial z}{\partial x}(1) = \frac{-19.240498}{0.132665} \] \[ \frac{\partial z}{\partial x}(1) \approx -144.96 \] So the derivative of \( z \) with respect to \( x \) at \( x = 1 \) is approximately \( -144.96 \). |
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| wenxinyiyan | 为了求解这个问题，我们首先需要找到函数$z$关于$x$的导数。给定函数为： $$z = \frac{1 + e^{-2x}}{x + \tan(12x)}$$ 我们可以使用商的导数规则来找到$z$关于$x$的导数。商的导数规则是： $$\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$$ 其中$u$和$v$是关于$x$的可导函数。在我们的情况下，令$u = 1 + e^{-2x}$和$v = x + \tan(12x)$。 首先，我们找到$u'$和$v'$： $$u' = \frac{d}{dx}(1 + e^{-2x}) = -2e^{-2x}$$ $$v' = \frac{d}{dx}(x + \tan(12x)) = 1 + 12\sec^2(12x)$$ 现在我们可以使用商的导数规则来找到$\frac{\partial z}{\partial x}$： $$\frac{\partial z}{\partial x} = \frac{(-2e^{-2x})(x + \tan(12x)) - (1 + e^{-2x})(1 + 12\sec^2(12x))}{(x + \tan(12x))^2}$$ 最后，为了找到$\frac{\partial z}{\partial x}$在$x = 1$时的值，我们将$x = 1$代入上述表达式中： $$\frac{\partial z}{\partial x}\Bigg|_{x=1} = \frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12\sec^2(12))}{(1 + \tan(12))^2}$$ 这个表达式给出了$\frac{\partial z}{\partial x}$在$x = 1$时的具体数值。注意，由于$\tan(12)$和$\sec(12)$的值可能非常大或非常小，因此在实际计算时可能需要使用数值方法来避免溢出或下溢。 请注意，由于$\tan(12)$在标准角度制下是一个非常大的值，这可能导致数值不稳定。在实际应用中，可以考虑使用弧度制，并对大数值进行适当处理，或者使用数学软件来进行精确计算。 另外，如果题目中的$\tan(12x)$是指$\tan(12) \times x$，那么需要对上述解答过程进行相应的调整。但从题目的表述来看，我们理解为$\tan$函数的参数是$12x$。 |
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| zhipu | To find the derivative of \( z \) with respect to \( x \), denoted as \( \frac{\partial z}{\partial x} \), we need to apply the quotient rule since \( z \) is a quotient of two functions. The quotient rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), its derivative is given by: \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \] For our function \( z = \frac{1 + e^{-2x}}{x + \tan(12x)} \), let: - \( g(x) = 1 + e^{-2x} \) - \( h(x) = x + \tan(12x) \) First, we need to find the derivatives \( g'(x) \) and \( h'(x) \): 1. **Derivative of \( g(x) \):** \[ g(x) = 1 + e^{-2x} \] Using the chain rule, the derivative is: \[ g'(x) = 0 + (-2)e^{-2x} = -2e^{-2x} \] 2. **Derivative of \( h(x) \):** \[ h(x) = x + \tan(12x) \] Using the sum rule and the chain rule, the derivative is: \[ h'(x) = 1 + 12 \sec^2(12x) \] Now, applying the quotient rule: \[ \frac{\partial z}{\partial x} = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \] Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \): \[ \frac{\partial z}{\partial x} = \frac{(-2e^{-2x})(x + \tan(12x)) - (1 + e^{-2x})(1 + 12 \sec^2(12x))}{(x + \tan(12x))^2} \] Now, we need to evaluate this derivative at \( x = 1 \): 1. **Evaluate \( g(1) \):** \[ g(1) = 1 + e^{-2 \cdot 1} = 1 + e^{-2} \] 2. **Evaluate \( g'(1) \):** \[ g'(1) = -2e^{-2 \cdot 1} = -2e^{-2} \] 3. **Evaluate \( h(1) \):** \[ h(1) = 1 + \tan(12 \cdot 1) = 1 + \tan(12) \] 4. **Evaluate \( h'(1) \):** \[ h'(1) = 1 + 12 \sec^2(12 \cdot 1) = 1 + 12 \sec^2(12) \] Substitute these values into the derivative expression: \[ \frac{\partial z}{\partial x} \bigg|_{x=1} = \frac{(-2e^{-2})(1 + \tan(12)) - (1 + e^{-2})(1 + 12 \sec^2(12))}{(1 + \tan(12))^2} \] Now, calculate the numerator: \[ (-2e^{-2})(1 + \tan(12)) = -2e^{-2} - 2e^{-2}\tan(12) \] \[ (1 + e^{-2})(1 + 12 \sec^2(12)) = 1 + 12 \sec^2(12) + e^{-2} + 12e^{-2} \sec^2(12) \] So the numerator becomes: \[ -2e^{-2} - 2e^{-2}\tan(12) - (1 + 12 \sec^2(12) + e^{-2} + 12e^{-2} \sec^2(12)) \] Combine like terms: \[ -2e^{-2} - 2e^{-2}\tan(12) - 1 - 12 \sec^2(12) - e^{-2} - 12e^{-2} \sec^2(12) \] \[ = -3e^{-2} - 2e^{-2}\tan(12) - 12e^{-2} \sec^2(12) - 1 - 12 \sec^2(12) \] The denominator is: \[ (1 + \tan(12))^2 \] Thus, the final expression for the derivative at \( x = 1 \) is: \[ \frac{\partial z}{\partial x} \bigg|_{x=1} = \frac{-3e^{-2} - 2e^{-2}\tan(12) - 12e^{-2} \sec^2(12) - 1 - 12 \sec^2(12)}{(1 + \tan(12))^2} \] This is the derivative of \( z \) with respect to \( x \) evaluated at \( x = 1 \). |
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| gpto1 | To find the derivative \(\frac{\partial z}{\partial x}\) of the function \[ z = \frac{1 + e^{-2x}}{x + \tan(12x)} \] at \( x = 1 \), we can follow these steps: **Step 1: Identify and simplify the numerator and denominator functions.** Let’s denote: - \( N(x) = 1 + e^{-2x} \) (the numerator), - \( D(x) = x + \tan(12x) \) (the denominator). Then: \[ z = \frac{N(x)}{D(x)} \] **Step 2: Compute the derivatives of the numerator and denominator.** Compute \( N'(x) \) and \( D'(x) \): 1. \( N'(x) = \frac{d}{dx}[1 + e^{-2x}] = -2e^{-2x} \). 2. \( D'(x) = \frac{d}{dx}[x + \tan(12x)] = 1 + 12\sec^2(12x) \). **Step 3: Use the quotient rule to find \( \frac{dz}{dx} \).** The quotient rule states: \[ \frac{d}{dx}\left(\frac{N(x)}{D(x)}\right) = \frac{N'(x)D(x) - N(x)D'(x)}{[D(x)]^2} \] So: \[ \frac{dz}{dx} = \frac{[-2e^{-2x}][x + \tan(12x)] - [1 + e^{-2x}][1 + 12\sec^2(12x)]}{[x + \tan(12x)]^2} \] **Step 4: Evaluate the numerator and denominator at \( x = 1 \).** We need to compute the values of \( N(1) \), \( N'(1) \), \( D(1) \), and \( D'(1) \). 1. Compute \( e^{-2(1)} = e^{-2} \approx 0.1353 \). 2. Then: - \( N(1) = 1 + e^{-2} \approx 1 + 0.1353 = 1.1353 \). - \( N'(1) = -2e^{-2} \approx -2 \times 0.1353 = -0.2707 \). 3. Compute \( \tan(12(1)) = \tan(12) \) and \( \sec^2(12(1)) = \sec^2(12) \). - Since 12 radians is not a standard angle, we'll compute these numerically. 4. Adjusting the angle: - \( 12 \) radians is equivalent to \( 12 \mod \pi \) because the tangent function has a period of \( \pi \). - \( \frac{12}{\pi} \approx 3.8197 \), so \( 12 - 3\pi \approx 0.8197\pi \). - Therefore, \( \tan(12) = \tan(0.8197\pi) \). 5. Compute \( \tan(0.8197\pi) \): - \( 0.8197\pi \) corresponds to \( 147.55^\circ \). - \( \tan(147.55^\circ) = -\tan(32.45^\circ) \approx -0.6357 \). 6. Compute \( \sec^2(12) \): - \( \cos(147.55^\circ) \approx -0.8401 \). - \( \sec(12) = \frac{1}{\cos(12)} = \frac{1}{-0.8401} = -1.1904 \). - \( \sec^2(12) = [-1.1904]^2 = 1.4171 \). 7. Then: - \( D(1) = 1 + \tan(12) \approx 1 + (-0.6357) = 0.3643 \). - \( D'(1) = 1 + 12\sec^2(12) \approx 1 + 12 \times 1.4171 = 1 + 17.0052 = 18.0052 \). **Step 5: Compute the numerator and denominator of the derivative at \( x = 1 \).** Compute: \[ \text{Numerator} = N'(1)D(1) - N(1)D'(1) = (-0.2707)(0.3643) - (1.1353)(18.0052) \] \[ \text{Numerator} \approx -0.0986 - 20.4325 = -20.5311 \] \[ \text{Denominator} = [D(1)]^2 = [0.3643]^2 = 0.1327 \] **Step 6: Compute \( \frac{dz}{dx} \) at \( x = 1 \).** \[ \frac{dz}{dx} \Big|_{x=1} = \frac{\text{Numerator}}{\text{Denominator}} = \frac{-20.5311}{0.1327} \] \[ \frac{dz}{dx} \Big|_{x=1} \approx -154.7783 \] **Answer:** Approximately \(-154.78\); that is, \(\dfrac{\partial z}{\partial x}\bigg|_{x=1} \approx -154.78\) |
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| deepseekv3 | To find the derivative \(\frac{\partial z}{\partial x}\) at \(x = 1\) for the function: \[ z = \frac{1 + e^{-2x}}{x + \tan(12x)}, \] we need to differentiate \(z\) with respect to \(x\) using the **quotient rule**. The quotient rule states that if \(z = \frac{u}{v}\), then: \[ \frac{dz}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}. \] ### Step 1: Identify \(u\) and \(v\) Here, \(u = 1 + e^{-2x}\) and \(v = x + \tan(12x)\). ### Step 2: Compute \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) - For \(u = 1 + e^{-2x}\), the derivative is: \[ \frac{du}{dx} = 0 + (-2)e^{-2x} = -2e^{-2x}. \] - For \(v = x + \tan(12x)\), the derivative is: \[ \frac{dv}{dx} = 1 + 12 \sec^2(12x). \] ### Step 3: Apply the quotient rule Substitute \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the quotient rule: \[ \frac{\partial z}{\partial x} = \frac{(x + \tan(12x))(-2e^{-2x}) - (1 + e^{-2x})(1 + 12 \sec^2(12x))}{(x + \tan(12x))^2}. \] ### Step 4: Evaluate at \(x = 1\) Substitute \(x = 1\) into the expression: \[ \frac{\partial z}{\partial x}\bigg|_{x=1} = \frac{(1 + \tan(12))(-2e^{-2}) - (1 + e^{-2})(1 + 12 \sec^2(12))}{(1 + \tan(12))^2}. \] ### Step 5: Simplify Since \(\tan(12)\) and \(\sec(12)\) are constants, the expression can be evaluated numerically: \[ \frac{\partial z}{\partial x}\bigg|_{x=1} \approx \frac{(1 + \tan(12))(-2e^{-2}) - (1 + e^{-2})(1 + 12 \sec^2(12))}{(1 + \tan(12))^2}. \] This is the final result. You can compute the numerical value using a calculator. |
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