| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the derivative of \( z = \arctan(e^{1 + (1 + x)^2}) \) with respect to \( x \), we will use the chain rule. The chain rule states that if \( z = f(g(x)) \), then \( \frac{dz}{dx} = f'(g(x)) \cdot g'(x) \). Let's apply this to our function: Let \( u = e^{1 + (1 + x)^2} \), then \( z = \arctan(u) \). We need to find \( \frac{du}{dx} \) and \( \frac{dz}{du} \). 1. \( \frac{dz}{du} = \frac{d}{du} \arctan(u) = \frac{1}{1+u^2} \) 2. \( \frac{du}{dx} = \frac{d}{dx} e^{1 + (1 + x)^2} \) Let's break down \( \frac{du}{dx} \) further: Let \( v = 1 + (1 + x)^2 \), then \( u = e^v \). Now, we need to find \( \frac{dv}{dx} \) and \( \frac{du}{dv} \). 1. \( \frac{du}{dv} = \frac{d}{dv} e^v = e^v = e^{1 + (1 + x)^2} \) 2. \( \frac{dv}{dx} = \frac{d}{dx} (1 + (1 + x)^2) = 2(1 + x) \) Now, we can find \( \frac{du}{dx} \) using the chain rule: \( \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = e^{1 + (1 + x)^2} \cdot 2(1 + x) \) Finally, we can find \( \frac{dz}{dx} \) using the chain rule: \( \frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} = \frac{1}{1+u^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \) Substituting \( u = e^{1 + (1 + x)^2} \): \( \frac{dz}{dx} = \frac{1}{1+e^{2(1 + (1 + x)^2)}} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \) Now we need to find the derivative at \( x = 0 \): \( \frac{dz}{dx}\bigg|_{x=0} = \frac{1}{1+e^{2(1 + (1)^2)}} \cdot e^{1 + (1)^2} \cdot 2(1) \) \( \frac{dz}{dx}\bigg|_{x=0} = \frac{1}{1+e^{4}} \cdot e^{2} \cdot 2 \) So, the derivative of \( z \) with respect to \( x \) at \( x = 0 \) is \( \frac{1}{1+e^{4}} \cdot e^{2} \cdot 2 \). |
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| deepseekr1 | To find the derivative of \( z = \arctan(e^{1 + (1 + x)^2}) \) with respect to \( x \) at \( x = 0 \), we use the chain rule multiple times. 1. Let \( u = e^{1 + (1 + x)^2} \). Then \( z = \arctan(u) \). 2. The derivative of \( \arctan(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \). 3. The derivative of \( u = e^{v} \) with respect to \( v \) (where \( v = 1 + (1 + x)^2 \)) is \( e^v \). 4. The derivative of \( v = 1 + (1 + x)^2 \) with respect to \( x \) is \( 2(1 + x) \). Applying the chain rule: \[ \frac{\partial z}{\partial x} = \frac{1}{1 + u^2} \cdot e^v \cdot 2(1 + x) \] Substituting \( u = e^v \) back into the expression: \[ \frac{\partial z}{\partial x} = \frac{e^v}{1 + e^{2v}} \cdot 2(1 + x) \] Evaluating at \( x = 0 \): - \( v = 1 + (1 + 0)^2 = 2 \) - \( e^v = e^2 \) - \( 1 + e^{2v} = 1 + e^4 \) - \( 2(1 + x) = 2 \cdot 1 = 2 \) Substituting these values into the derivative: \[ \frac{\partial z}{\partial x} \bigg|_{x=0} = \frac{e^2}{1 + e^4} \cdot 2 = \frac{2e^2}{1 + e^4} \] Thus, the derivative of \(\frac{\partial z}{\partial x}\) at \(x = 0\) is \boxed{\dfrac{2e^2}{1 + e^4}}. |
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| gpto1 | To find the derivative \(\frac{\partial z}{\partial x}\) at \(x = 0\) for the function \[ z = \arctan\left( e^{1 + (1 + x)^2} \right), \] we will apply the chain rule and compute the derivative step by step. **Step 1: Identify the inner function and compute its derivative.** Let \[ u(x) = e^{1 + (1 + x)^2}. \] Then \[ z = \arctan(u(x)). \] First, compute \( u'(x) = \frac{du}{dx} \). Since \( u(x) = e^{v(x)} \), where \( v(x) = 1 + (1 + x)^2 \), we have \[ u'(x) = e^{v(x)} \cdot v'(x). \] Compute \( v'(x) \): \[ v(x) = 1 + (1 + x)^2, \] \[ v'(x) = 2(1 + x) \cdot (1) = 2(1 + x). \] So, \[ u'(x) = e^{1 + (1 + x)^2} \cdot 2(1 + x). \] **Step 2: Compute the derivative of the outer function.** The derivative of \( z = \arctan(u) \) with respect to \( u \) is: \[ \frac{dz}{du} = \frac{1}{1 + u^2}. \] **Step 3: Apply the chain rule to find \( \frac{dz}{dx} \).** \[ \frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot u'(x). \] Substitute \( u(x) \) and \( u'(x) \): \[ \frac{dz}{dx} = \frac{1}{1 + \left( e^{1 + (1 + x)^2} \right)^2} \cdot \left[ e^{1 + (1 + x)^2} \cdot 2(1 + x) \right]. \] Simplify: \[ \frac{dz}{dx} = \frac{2(1 + x)e^{1 + (1 + x)^2}}{1 + e^{2[1 + (1 + x)^2]}}. \] **Step 4: Evaluate \( \frac{dz}{dx} \) at \( x = 0 \).** First, compute \( e^{1 + (1 + x)^2} \) at \( x = 0 \): \[ (1 + x) = 1 + 0 = 1, \] \[ (1 + x)^2 = 1^2 = 1, \] \[ e^{1 + (1 + x)^2} = e^{1 + 1} = e^{2}. \] Next, compute the denominator: \[ 1 + e^{2[1 + (1 + x)^2]} = 1 + e^{2 \cdot 2} = 1 + e^{4}. \] Now, substitute back into \( \frac{dz}{dx} \): \[ \frac{dz}{dx}\Big|_{x = 0} = \frac{2(1)(e^{2})}{1 + e^{4}} = \frac{2e^{2}}{1 + e^{4}}. \] **Final Answer:** \[ \frac{\partial z}{\partial x}\Bigg|_{x=0} \;=\; \frac{2\,e^{2}}{1\,+\,e^{4}}. \] **Answer:** An explicit expression: (2 × e²) divided by (1 + e⁴); that is, ∂z/∂x at x = 0 = [2e²]⁄[1 + e⁴] |
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| GPT-4o | To find the derivative of \( z = \arctan(e^{1 + (1 + x)^2}) \) with respect to \( x \) at \( x = 0 \), we will follow these steps: 1. **Express \( z \) and find its first derivative with respect to \( x \)**: \[ z = \arctan\left(e^{1 + (1 + x)^2}\right) \] 2. **Let \( u = e^{1 + (1 + x)^2} \)**. Then: \[ z = \arctan(u) \] 3. **Find \( \frac{dz}{dx} \) using the chain rule**: \[ \frac{dz}{dx} = \frac{d}{dx} \left( \arctan(u) \right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] 4. **Compute \( \frac{du}{dx} \)**. Recall \( u = e^{1 + (1 + x)^2} \): \[ \frac{du}{dx} = e^{1 + (1 + x)^2} \cdot \frac{d}{dx} \left( 1 + (1 + x)^2 \right) \] \[ \frac{d}{dx} \left( 1 + (1 + x)^2 \right) = \frac{d}{dx} \left( 1 + 1 + 2x + x^2 \right) = 2x + 2 \] \[ \frac{du}{dx} = e^{1 + (1 + x)^2} \cdot (2x + 2) \] 5. **Combine these results**: \[ \frac{dz}{dx} = \frac{1}{1 + \left(e^{1 + (1 + x)^2}\right)^2} \cdot e^{1 + (1 + x)^2} \cdot (2x + 2) \] 6. **Evaluate this expression at \( x = 0 \)**: \[ u \bigg|_{x=0} = e^{1 + (1 + 0)^2} = e^2 \] \[ \left. \frac{dz}{dx} \right|_{x=0} = \frac{1}{1 + (e^2)^2} \cdot e^2 \cdot (2(0) + 2) \] \[ \left. \frac{dz}{dx} \right|_{x=0} = \frac{1}{1 + e^4} \cdot e^2 \cdot 2 \] \[ \left. \frac{dz}{dx} \right|_{x=0} = \frac{2e^2}{1 + e^4} \] Thus, the derivative of \( z \) with respect to \( x \) at \( x = 0 \) is: \[ \left. \frac{\partial z}{\partial x} \right|_{x=0} = \frac{2e^2}{1 + e^4} \] |
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| llama3.1-70B-Instruct | To find the derivative of \( z = \arctan(e^{1 + (1 + x)^2}) \) with respect to \( x \) at \( x = 0 \), we'll apply the chain rule. The derivative of \( \arctan(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \), and the derivative of \( e^u \) with respect to \( u \) is \( e^u \). We'll also need the derivative of \( 1 + (1 + x)^2 \) with respect to \( x \), which is \( 2(1 + x) \). Given: \[ z = \arctan(e^{1 + (1 + x)^2}) \] Let's denote \( u = e^{1 + (1 + x)^2} \). Then: \[ \frac{\partial z}{\partial u} = \frac{1}{1 + u^2} \] And: \[ \frac{\partial u}{\partial x} = e^{1 + (1 + x)^2} \cdot 2(1 + x) \] By the chain rule: \[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} \] Substituting the expressions for \( \frac{\partial z}{\partial u} \) and \( \frac{\partial u}{\partial x} \): \[ \frac{\partial z}{\partial x} = \frac{1}{1 + (e^{1 + (1 + x)^2})^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \] Simplifying: \[ \frac{\partial z}{\partial x} = \frac{2(1 + x)e^{1 + (1 + x)^2}}{1 + e^{2(1 + (1 + x)^2)}} \] To find the derivative at \( x = 0 \), we substitute \( x = 0 \) into the expression: \[ \frac{\partial z}{\partial x} \Big|_{x = 0} = \frac{2(1 + 0)e^{1 + (1 + 0)^2}}{1 + e^{2(1 + (1 + 0)^2)}} \] \[ \frac{\partial z}{\partial x} \Big|_{x = 0} = \frac{2e^2}{1 + e^4} \] This is the derivative of \( z \) with respect to \( x \) at \( x = 0 \). |
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| llama3.1-405B-Instruct | To find the derivative of z with respect to x, we can use the chain rule and the fact that the derivative of arctan(u) is 1/(1+u^2). First, let's rewrite the expression for z: z = arctan(e^(1 + (1 + x)^2)) Now, let's find the derivative of z with respect to x: dz/dx = d(arctan(e^(1 + (1 + x)^2)))/dx Using the chain rule, we can write: dz/dx = (1/(1+e^(1 + (1 + x)^2))) \* d(e^(1 + (1 + x)^2))/dx Now, let's find the derivative of e^(1 + (1 + x)^2) with respect to x: d(e^(1 + (1 + x)^2))/dx = e^(1 + (1 + x)^2) \* d(1 + (1 + x)^2)/dx Using the chain rule again, we can write: d(1 + (1 + x)^2)/dx = 2(1 + x) So, we have: d(e^(1 + (1 + x)^2))/dx = e^(1 + (1 + x)^2) \* 2(1 + x) Now, we can substitute this expression back into the equation for dz/dx: dz/dx = (1/(1+e^(1 + (1 + x)^2))) \* e^(1 + (1 + x)^2) \* 2(1 + x) Simplifying, we get: dz/dx = 2(1 + x)/(1+e^(1 + (1 + x)^2)) Now, we need to find the value of this derivative at x = 0: dz/dx(0) = 2(1 + 0)/(1+e^(1 + (1 + 0)^2)) Simplifying, we get: dz/dx(0) = 2/(1+e^2) So, the derivative of z with respect to x at x = 0 is 2/(1+e^2). |
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| DeepSeek-Coder-V2 | To solve the problem of finding the derivative of \( \frac{\partial z}{\partial x} \) at \( x = 0 \) for the function \( z = \arctan(e^{1 + (1 + x)^2}) \), we will follow these steps: 1. **Express \( z \) in terms of \( x \):** \[ z = \arctan(e^{1 + (1 + x)^2}) \] 2. **Differentiate \( z \) with respect to \( x \):** We use the chain rule to differentiate \( z \). Let \( u = e^{1 + (1 + x)^2} \). Then, \( z = \arctan(u) \). The derivative of \( z \) with respect to \( u \) is: \[ \frac{dz}{du} = \frac{1}{1 + u^2} \] Next, we need to find \( \frac{du}{dx} \). Let \( v = 1 + (1 + x)^2 \). Then, \( u = e^v \). The derivative of \( u \) with respect to \( v \) is: \[ \frac{du}{dv} = e^v \] Now, we need to find \( \frac{dv}{dx} \). We have: \[ v = 1 + (1 + x)^2 \] Differentiate \( v \) with respect to \( x \): \[ \frac{dv}{dx} = 2(1 + x) \] Combining these results, we get: \[ \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = e^v \cdot 2(1 + x) \] Substitute \( v = 1 + (1 + x)^2 \) back into the expression: \[ \frac{du}{dx} = e^{1 + (1 + x)^2} \cdot 2(1 + x) \] 3. **Combine the derivatives:** Using the chain rule, we have: \[ \frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \] Substitute \( u = e^{1 + (1 + x)^2} \) back into the expression: \[ \frac{dz}{dx} = \frac{1}{1 + (e^{1 + (1 + x)^2})^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \] 4. **Evaluate the derivative at \( x = 0 \):** Substitute \( x = 0 \) into the expression: \[ \frac{dz}{dx} \bigg|_{x=0} = \frac{1}{1 + (e^{1 + (1 + 0)^2})^2} \cdot e^{1 + (1 + 0)^2} \cdot 2(1 + 0) \] Simplify the expression: \[ \frac{dz}{dx} \bigg|_{x=0} = \frac{1}{1 + e^{1 + 1^2}} \cdot e^{1 + 1^2} \cdot 2 \] \[ \frac{dz}{dx} \bigg|_{x=0} = \frac{1}{1 + e^2} \cdot e^2 \cdot 2 \] \[ \frac{dz}{dx} \bigg|_{x=0} = \frac{2e^2}{1 + e^2} \] Therefore, the derivative of \( \frac{\partial z}{\partial x} \) at \( x = 0 \) is: \[ \boxed{\frac{2e^2}{1 + e^2}} \] |
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| Qwen2-72B-Instruct | To find the derivative of $z = \arctan(e^{1 + (1 + x)^2})$ with respect to $x$, we will use the chain rule. The chain rule states that if $z = f(g(x))$, then $\frac{dz}{dx} = f'(g(x)) \cdot g'(x)$. Let's apply this to our function: Let $u = e^{1 + (1 + x)^2}$, then $z = \arctan(u)$. We need to find $\frac{du}{dx}$ and $\frac{dz}{du}$. 1. $\frac{dz}{du} = \frac{d}{du} \arctan(u) = \frac{1}{1+u^2}$ 2. $\frac{du}{dx} = \frac{d}{dx} e^{1 + (1 + x)^2}$ Let's break down $\frac{du}{dx}$ further: Let $v = 1 + (1 + x)^2$, then $u = e^v$. Now, we need to find $\frac{dv}{dx}$ and $\frac{du}{dv}$. 1. $\frac{du}{dv} = \frac{d}{dv} e^v = e^v = e^{1 + (1 + x)^2}$ 2. $\frac{dv}{dx} = \frac{d}{dx} (1 + (1 + x)^2) = 2(1 + x)$ Now, we can find $\frac{du}{dx}$ using the chain rule: $\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = e^{1 + (1 + x)^2} \cdot 2(1 + x)$ Finally, we can find $\frac{dz}{dx}$ using the chain rule: $\frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} = \frac{1}{1+u^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x)$ Substituting $u = e^{1 + (1 + x)^2}$: $\frac{dz}{dx} = \frac{1}{1+e^{2(1 + (1 + x)^2)}} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x)$ Now we need to find the derivative at $x = 0$: $\frac{dz}{dx}\bigg|_{x=0} = \frac{1}{1+e^{2(1 + (1)^2)}} \cdot e^{1 + (1)^2} \cdot 2(1)$ $\frac{dz}{dx}\bigg|_{x=0} = \frac{1}{1+e^{4}} \cdot e^{2} \cdot 2$ So, the derivative of $z$ with respect to $x$ at $x = 0$ is $\frac{1}{1+e^{4}} \cdot e^{2} \cdot 2$. |
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| baichuan4 | To find the derivative $\frac{\partial z}{\partial x}$ of $z = \arctan(e^{1 + (1 + x)^2})$, we can use the chain rule. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. In our case, we have $z = \arctan(u)$, where $u = e^{1 + (1 + x)^2}$. We need to find the derivative of $z$ with respect to $x$, which involves finding the derivative of $u$ with respect to $x$ as well. First, let's find $u'(x)$: $$u = e^{1 + (1 + x)^2}$$ $$u' = e^{1 + (1 + x)^2} \cdot 2(1 + x) \cdot 1$$ $$u' = 2(1 + x)e^{1 + (1 + x)^2}$$ Now, we know that the derivative of $\arctan(u)$ with respect to $u$ is $\frac{1}{1 + u^2}$. Therefore, applying the chain rule, we get: $$\frac{\partial z}{\partial x} = \frac{1}{1 + (e^{1 + (1 + x)^2})^2} \cdot 2(1 + x)e^{1 + (1 + x)^2}$$ $$\frac{\partial z}{\partial x} = \frac{2(1 + x)e^{1 + (1 + x)^2}}{1 + e^{2 + 2(1 + x)^2}}$$ Now, we need to evaluate this expression at $x = 0$: $$\frac{\partial z}{\partial x}\Big|_{x=0} = \frac{2(1 + 0)e^{1 + (1 + 0)^2}}{1 + e^{2 + 2(1 + 0)^2}}$$ $$\frac{\partial z}{\partial x}\Big|_{x=0} = \frac{2e^2}{1 + e^4}$$ This is the value of the derivative of $z$ with respect to $x$ at $x = 0$. |
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| doubao | Let $u=1+(1+x)^2$. Then $\frac{du}{dx}=2(1+x)$. When $x=0$, $u=1+(1+0)^2=1+1=2$. Now $z=\arctan(e^u)$. The derivative of $\arctan t$ is $\frac{1}{1+t^2}$. So $\frac{dz}{du}=\frac{1}{1+e^{2u}}$. And the derivative of $e^t$ is $e^t$. So $\frac{dz}{dx}=\frac{dz}{du}\cdot\frac{du}{dx}=\frac{1}{1+e^{2u}}\cdot 2(1+x)$. When $x=0$ and $u=2$, $\frac{\partial z}{\partial x}=\frac{1}{1+e^{4}}\cdot 2(1+0)=\frac{2}{1+e^{4}}$. |
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| minimax | To find the derivative of \( \frac{\partial z}{\partial x} \) at \( x = 0 \), we need to differentiate \( z \) with respect to \( x \). Given that \( z = \arctan(e^{1 + (1 + x)^2}) \), we can use the chain rule to differentiate this function. Let's denote \( u = 1 + (1 + x)^2 \) and \( v = e^u \). Then \( z = \arctan(v) \). First, we differentiate \( u \) with respect to \( x \): \( u = 1 + (1 + x)^2 \) \( u' = 2(1 + x) \) At \( x = 0 \), \( u' = 2(1 + 0) = 2 \). Next, we differentiate \( v \) with respect to \( u \): \( v = e^u \) \( \frac{dv}{du} = e^u \) Now, we differentiate \( z \) with respect to \( v \): \( z = \arctan(v) \) \( \frac{dz}{dv} = \frac{1}{1 + v^2} \) Using the chain rule, the derivative of \( z \) with respect to \( x \) is: \( \frac{dz}{dx} = \frac{dz}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} \) Substituting the derivatives we found: \( \frac{dz}{dx} = \frac{1}{1 + v^2} \cdot e^u \cdot 2 \) Now, we substitute \( v = e^u \) and \( u = 1 + (1 + x)^2 \) back into the equation: \( \frac{dz}{dx} = \frac{1}{1 + (e^u)^2} \cdot e^u \cdot 2 \) At \( x = 0 \), \( u = 1 + (1 + 0)^2 = 1 + 1 = 2 \), so \( v = e^u = e^2 \). \( \frac{dz}{dx} \bigg|_{x=0} = \frac{1}{1 + (e^2)^2} \cdot e^2 \cdot 2 \) \( \frac{dz}{dx} \bigg|_{x=0} = \frac{2e^2}{1 + e^4} \) This is the derivative of \( z \) with respect to \( x \) evaluated at \( x = 0 \). |
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| kimi | To find the derivative of \( z \) with respect to \( x \), we'll use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Given: \[ z = \arctan(e^{1 + (1 + x)^2}) \] Let's denote the inner function as \( u \), so: \[ u = e^{1 + (1 + x)^2} \] Then \( z = \arctan(u) \). Now, we'll find the derivatives of the outer and inner functions. The derivative of \( \arctan(u) \) with respect to \( u \) is: \[ \frac{d}{du}(\arctan(u)) = \frac{1}{1 + u^2} \] The derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = \frac{d}{dx}(e^{1 + (1 + x)^2}) \] \[ \frac{du}{dx} = e^{1 + (1 + x)^2} \cdot \frac{d}{dx}(1 + (1 + x)^2) \] \[ \frac{du}{dx} = e^{1 + (1 + x)^2} \cdot 2(1 + x) \] Now, applying the chain rule: \[ \frac{\partial z}{\partial x} = \frac{d}{du}(\arctan(u)) \cdot \frac{du}{dx} \] \[ \frac{\partial z}{\partial x} = \frac{1}{1 + u^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \] Substitute \( u = e^{1 + (1 + x)^2} \) into the equation: \[ \frac{\partial z}{\partial x} = \frac{1}{1 + (e^{1 + (1 + x)^2})^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \] \[ \frac{\partial z}{\partial x} = \frac{e^{1 + (1 + x)^2}}{1 + e^{2 + 2(1 + x)^2}} \cdot 2(1 + x) \] Now, we need to find the value of this derivative at \( x = 0 \): \[ \frac{\partial z}{\partial x} \bigg|_{x=0} = \frac{e^{1 + (1 + 0)^2}}{1 + e^{2 + 2(1 + 0)^2}} \cdot 2(1 + 0) \] \[ \frac{\partial z}{\partial x} \bigg|_{x=0} = \frac{e^2}{1 + e^2} \cdot 2 \] \[ \frac{\partial z}{\partial x} \bigg|_{x=0} = \frac{2e^2}{1 + e^2} \] So, the derivative of \( z \) with respect to \( x \) at \( x = 0 \) is \( \frac{2e^2}{1 + e^2} \). |
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| wenxinyiyan | 为了求解这个问题，我们首先需要找到$z$关于$x$的偏导数。给定 $z = \arctan(e^{1 + (1 + x)^2})$，我们可以使用链式法则来求导。 步骤1: 对$\arctan(u)$求导，其中$u = e^{1 + (1 + x)^2}$。 $\frac{d}{du}(\arctan(u)) = \frac{1}{1 + u^2}$步骤2: 对$e^{1 + (1 + x)^2}$求导。 令$v = 1 + (1 + x)^2$，则$\frac{dv}{dx} = 2(1+x)$。 $\frac{d}{dv}(e^v) = e^v \cdot \frac{dv}{dx} = e^{1 + (1 + x)^2} \cdot 2(1+x)$步骤3: 应用链式法则求得$\frac{\partial z}{\partial x}$。 $\frac{\partial z}{\partial x} = \frac{1}{1 + (e^{1 + (1 + x)^2})^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1+x)$步骤4: 将$x=0$代入$\frac{\partial z}{\partial x}$。 $\frac{\partial z}{\partial x}\bigg|_{x=0} = \frac{1}{1 + (e^{1 + (1 + 0)^2})^2} \cdot e^{1 + (1 + 0)^2} \cdot 2(1+0)$$= \frac{1}{1 + e^4} \cdot e^2 \cdot 2$$= \frac{2e^2}{1 + e^4}$所以，当$x=0$时，$\frac{\partial z}{\partial x} = \frac{2e^2}{1 + e^4}$。 |
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| zhipu | To find the derivative of \( z = \arctan(e^{1 + (1 + x)^2}) \) with respect to \( x \), we will use the chain rule. Let's break it down step by step. 1. **Identify the outer and inner functions:** - The outer function is \( \arctan(u) \), where \( u = e^{1 + (1 + x)^2} \). - The inner function is \( u = e^{1 + (1 + x)^2} \). 2. **Differentiate the outer function:** The derivative of \( \arctan(u) \) with respect to \( u \) is: \[ \frac{d}{du} \arctan(u) = \frac{1}{1 + u^2} \] 3. **Differentiate the inner function:** The inner function \( u = e^{1 + (1 + x)^2} \) can be differentiated with respect to \( x \) using the chain rule. First, let's differentiate the exponent \( 1 + (1 + x)^2 \): \[ \frac{d}{dx} [1 + (1 + x)^2] = \frac{d}{dx} [1 + 1 + 2x + x^2] = 2(1 + x) \] Now, differentiate \( e^{1 + (1 + x)^2} \): \[ \frac{d}{dx} e^{1 + (1 + x)^2} = e^{1 + (1 + x)^2} \cdot \frac{d}{dx} [1 + (1 + x)^2] = e^{1 + (1 + x)^2} \cdot 2(1 + x) \] 4. **Combine the derivatives using the chain rule:** \[ \frac{\partial z}{\partial x} = \frac{d}{du} \arctan(u) \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \] Substitute \( u = e^{1 + (1 + x)^2} \) back into the equation: \[ \frac{\partial z}{\partial x} = \frac{1}{1 + (e^{1 + (1 + x)^2})^2} \cdot e^{1 + (1 + x)^2} \cdot 2(1 + x) \] 5. **Evaluate at \( x = 0 \):** First, calculate \( u \) at \( x = 0 \): \[ u = e^{1 + (1 + 0)^2} = e^{1 + 1^2} = e^2 \] Now, substitute \( x = 0 \) and \( u = e^2 \) into the derivative: \[ \left. \frac{\partial z}{\partial x} \right|_{x=0} = \frac{1}{1 + (e^2)^2} \cdot e^2 \cdot 2(1 + 0) = \frac{1}{1 + e^4} \cdot e^2 \cdot 2 \] Simplify the expression: \[ \left. \frac{\partial z}{\partial x} \right|_{x=0} = \frac{2e^2}{1 + e^4} \] Therefore, the derivative of \( z \) with respect to \( x \) at \( x = 0 \) is: \[ \boxed{\frac{2e^2}{1 + e^4}} \] |
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| gpto1 | To find the derivative \(\frac{\partial z}{\partial x}\) at \(x = 0\) for the function \[ z = \arctan\left( e^{1 + (1 + x)^2} \right), \] we will apply the chain rule and compute the derivative step by step. **Step 1: Identify the inner function and compute its derivative.** Let \[ u(x) = e^{1 + (1 + x)^2}. \] Then \[ z = \arctan(u(x)). \] First, compute \( u'(x) = \frac{du}{dx} \). Since \( u(x) = e^{v(x)} \), where \( v(x) = 1 + (1 + x)^2 \), we have \[ u'(x) = e^{v(x)} \cdot v'(x). \] Compute \( v'(x) \): \[ v(x) = 1 + (1 + x)^2, \] \[ v'(x) = 2(1 + x) \cdot (1) = 2(1 + x). \] So, \[ u'(x) = e^{1 + (1 + x)^2} \cdot 2(1 + x). \] **Step 2: Compute the derivative of the outer function.** The derivative of \( z = \arctan(u) \) with respect to \( u \) is: \[ \frac{dz}{du} = \frac{1}{1 + u^2}. \] **Step 3: Apply the chain rule to find \( \frac{dz}{dx} \).** \[ \frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot u'(x). \] Substitute \( u(x) \) and \( u'(x) \): \[ \frac{dz}{dx} = \frac{1}{1 + \left( e^{1 + (1 + x)^2} \right)^2} \cdot \left[ e^{1 + (1 + x)^2} \cdot 2(1 + x) \right]. \] Simplify: \[ \frac{dz}{dx} = \frac{2(1 + x)e^{1 + (1 + x)^2}}{1 + e^{2[1 + (1 + x)^2]}}. \] **Step 4: Evaluate \( \frac{dz}{dx} \) at \( x = 0 \).** First, compute \( e^{1 + (1 + x)^2} \) at \( x = 0 \): \[ (1 + x) = 1 + 0 = 1, \] \[ (1 + x)^2 = 1^2 = 1, \] \[ e^{1 + (1 + x)^2} = e^{1 + 1} = e^{2}. \] Next, compute the denominator: \[ 1 + e^{2[1 + (1 + x)^2]} = 1 + e^{2 \cdot 2} = 1 + e^{4}. \] Now, substitute back into \( \frac{dz}{dx} \): \[ \frac{dz}{dx}\Big|_{x = 0} = \frac{2(1)(e^{2})}{1 + e^{4}} = \frac{2e^{2}}{1 + e^{4}}. \] **Final Answer:** \[ \frac{\partial z}{\partial x}\Bigg|_{x=0} \;=\; \frac{2\,e^{2}}{1\,+\,e^{4}}. \] **Answer:** An explicit expression: (2 × e²) divided by (1 + e⁴); that is, ∂z/∂x at x = 0 = [2e²]⁄[1 + e⁴] |
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| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-77b57b8a-0721-9e53-9cb4-6da8163f02e6', 'request_id': '77b57b8a-0721-9e53-9cb4-6da8163f02e6'} | 0.3017 |