| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the function \( e^{4 x^5-4} + \cos \left(2 x^4+8\right) \), we will use the chain rule for differentiation. Let's break it down step by step. First, consider the function \( e^{4 x^5-4} \): 1. Let \( u = 4x^5 - 4 \). 2. The derivative of \( e^u \) with respect to \( u \) is \( e^u \). 3. Now, we need to differentiate \( u = 4x^5 - 4 \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx}(4x^5 - 4) = 20x^4 \] 4. Using the chain rule, the derivative of \( e^{4 x^5-4} \) with respect to \( x \) is: \[ \frac{d}{dx} e^{4 x^5-4} = e^{4 x^5-4} \cdot 20x^4 \] Next, consider the function \( \cos \left(2 x^4+8\right) \): 1. Let \( v = 2x^4 + 8 \). 2. The derivative of \( \cos(v) \) with respect to \( v \) is \( -\sin(v) \). 3. Now, we need to differentiate \( v = 2x^4 + 8 \) with respect to \( x \): \[ \frac{dv}{dx} = \frac{d}{dx}(2x^4 + 8) = 8x^3 \] 4. Using the chain rule, the derivative of \( \cos \left(2 x^4+8\right) \) with respect to \( x \) is: \[ \frac{d}{dx} \cos \left(2 x^4+8\right) = -\sin \left(2 x^4+8\right) \cdot 8x^3 \] Combining the results from both parts, the derivative of the entire function \( e^{4 x^5-4} + \cos \left(2 x^4+8\right) \) is: \[ \frac{d}{dx} \left( e^{4 x^5-4} + \cos \left(2 x^4+8\right) \right) = 20x^4 e^{4 x^5-4} - 8x^3 \sin \left(2 x^4+8\right) \] Thus, the final answer is: \[ \boxed{20x^4 e^{4 x^5-4} - 8x^3 \sin \left(2 x^4+8\right)} \] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| deepseek-r1 | To differentiate the function \( e^{4 x^5 - 4} + \cos \left(2 x^4 + 8\right) \), we apply the chain rule to each term separately. 1. For the first term \( e^{4 x^5 - 4} \): - Let \( u = 4x^5 - 4 \). The derivative of \( e^u \) with respect to \( x \) is \( e^u \cdot u' \). - Compute \( u' \): \( \frac{d}{dx}(4x^5 - 4) = 20x^4 \). - Therefore, the derivative is \( e^{4x^5 - 4} \cdot 20x^4 \). 2. For the second term \( \cos(2x^4 + 8) \): - Let \( v = 2x^4 + 8 \). The derivative of \( \cos(v) \) with respect to \( x \) is \( -\sin(v) \cdot v' \). - Compute \( v' \): \( \frac{d}{dx}(2x^4 + 8) = 8x^3 \). - Therefore, the derivative is \( -\sin(2x^4 + 8) \cdot 8x^3 \). Combining the derivatives of both terms, we get: \[ \boxed{20x^4 e^{4x^5 - 4} - 8x^3 \sin\left(2x^4 + 8\right)} \] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| gpto1 | To find the derivative of the function: \[ f(x) = e^{4 x^5 - 4} + \cos(2 x^4 + 8), \] we'll differentiate each term separately. **First term:** \( e^{4 x^5 - 4} \) Let \( u = 4 x^5 - 4 \), so the derivative is: \[ \frac{d}{dx} e^{u} = e^{u} \cdot \frac{du}{dx} = e^{4 x^5 - 4} \cdot (20 x^4). \] **Second term:** \( \cos(2 x^4 + 8) \) Let \( v = 2 x^4 + 8 \), so the derivative is: \[ \frac{d}{dx} \cos(v) = -\sin(v) \cdot \frac{dv}{dx} = -\sin(2 x^4 + 8) \cdot (8 x^3). \] **Combining the derivatives:** \[ f'(x) = 20 x^4 e^{4 x^5 - 4} - 8 x^3 \sin(2 x^4 + 8). \] **Final Answer:** \[\boxed{f'(x)=20x^{4}e^{4x^{5}-4}\,-\,8x^{3}\sin(2x^{4}+8)}\] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| gpt-4o-2024-08-06 | To differentiate the function \( f(x) = e^{4x^5 - 4} + \cos(2x^4 + 8) \), we will apply the chain rule to each term separately. 1. **Differentiate \( e^{4x^5 - 4} \):** Let \( u = 4x^5 - 4 \). Then, the derivative of \( e^u \) with respect to \( x \) is given by: \[ \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx} \] First, find \( \frac{du}{dx} \): \[ \frac{du}{dx} = \frac{d}{dx}(4x^5 - 4) = 20x^4 \] Therefore, the derivative of \( e^{4x^5 - 4} \) is: \[ e^{4x^5 - 4} \cdot 20x^4 = 20x^4 e^{4x^5 - 4} \] 2. **Differentiate \( \cos(2x^4 + 8) \):** Let \( v = 2x^4 + 8 \). Then, the derivative of \( \cos(v) \) with respect to \( x \) is given by: \[ \frac{d}{dx} \cos(v) = -\sin(v) \cdot \frac{dv}{dx} \] First, find \( \frac{dv}{dx} \): \[ \frac{dv}{dx} = \frac{d}{dx}(2x^4 + 8) = 8x^3 \] Therefore, the derivative of \( \cos(2x^4 + 8) \) is: \[ -\sin(2x^4 + 8) \cdot 8x^3 = -8x^3 \sin(2x^4 + 8) \] 3. **Combine the derivatives:** The derivative of the entire function \( f(x) = e^{4x^5 - 4} + \cos(2x^4 + 8) \) is the sum of the derivatives of the individual terms: \[ \frac{d}{dx} f(x) = 20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8) \] Thus, the derivative of the function is: \[ \boxed{20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8)} \] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| DeepSeek-Coder-V2-Lite-Instruct | The differentiation of the function \( e^{4 x^5-4} + \cos \left(2 x^4+8\right) \) involves differentiating each term separately. 1. **Differentiating \( e^{4 x^5-4} \):** - Let \( u = 4 x^5 - 4 \). - The derivative of \( e^u \) with respect to \( u \) is \( e^u \). - The derivative of \( u \) with respect to \( x \) is \( \frac{d}{dx}(4 x^5 - 4) = 20 x^4 \). - Therefore, the derivative of \( e^{4 x^5-4} \) with respect to \( x \) is \( e^{4 x^5-4} \cdot 20 x^4 \). 2. **Differentiating \( \cos \left(2 x^4+8\right) \):** - Let \( v = 2 x^4 + 8 \). - The derivative of \( \cos v \) with respect to \( v \) is \( -\sin v \). - The derivative of \( v \) with respect to \( x \) is \( \frac{d}{dx}(2 x^4 + 8) = 8 x^3 \). - Therefore, the derivative of \( \cos \left(2 x^4+8\right) \) with respect to \( x \) is \( -\sin \left(2 x^4+8\right) \cdot 8 x^3 \). Combining these results, the derivative of the given function is: \[ \frac{d}{dx} \left( e^{4 x^5-4} + \cos \left(2 x^4+8\right) \right) = e^{4 x^5-4} \cdot 20 x^4 - \sin \left(2 x^4+8\right) \cdot 8 x^3 \] Thus, the final answer is: \[ \boxed{20 x^4 e^{4 x^5-4} - 8 x^3 \sin \left(2 x^4+8\right)} \] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the components of the function that need differentiation The given function is a sum of two components: $e^{4 x^5-4}$ and $\cos \left(2 x^4+8\right)$. Each of these components needs to be differentiated separately. ## Step 2: Differentiate the exponential component To differentiate $e^{4 x^5-4}$, we use the chain rule. The derivative of $e^u$ with respect to $u$ is $e^u$, and here $u = 4 x^5 - 4$. So, the derivative of this component is $e^{4 x^5-4} \cdot \frac{d}{dx}(4 x^5-4)$. Calculating $\frac{d}{dx}(4 x^5-4)$ gives $20 x^4$. ## Step 3: Differentiate the cosine component To differentiate $\cos \left(2 x^4+8\right)$, we again use the chain rule. The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$, and here $u = 2 x^4 + 8$. So, the derivative of this component is $-\sin \left(2 x^4+8\right) \cdot \frac{d}{dx}(2 x^4+8)$. Calculating $\frac{d}{dx}(2 x^4+8)$ gives $8 x^3$. ## Step 4: Combine the derivatives of the components Combining the results from steps 2 and 3, the derivative of the entire function is $e^{4 x^5-4} \cdot 20 x^4 - \sin \left(2 x^4+8\right) \cdot 8 x^3$. ## Step 5: Simplify the expression if necessary The expression $e^{4 x^5-4} \cdot 20 x^4 - \sin \left(2 x^4+8\right) \cdot 8 x^3$ is already simplified. The final answer is: $\boxed{20 x^4 e^{4 x^5-4} - 8 x^3 \sin \left(2 x^4+8\right)}$ |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the components of the function to be differentiated The given function is a sum of two functions: $e^{4 x^5-4}$ and $\cos \left(2 x^4+8\right)$. We will differentiate each component separately. ## Step 2: Differentiate the first component using the exponential derivative rule The derivative of $e^{u}$ with respect to $u$ is $e^{u}$. Here, $u = 4 x^5-4$, so we need to apply the composite function rule (chain rule). The derivative of $u$ with respect to $x$ is $20x^4$. Therefore, the derivative of $e^{4 x^5-4}$ is $e^{4 x^5-4} \cdot 20x^4 = 20x^4e^{4 x^5-4}$. ## Step 3: Differentiate the second component using the cosine derivative rule The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$. Here, $u = 2 x^4+8$, so we apply the composite function rule (chain rule) again. The derivative of $u$ with respect to $x$ is $8x^3$. Therefore, the derivative of $\cos \left(2 x^4+8\right)$ is $-\sin \left(2 x^4+8\right) \cdot 8x^3 = -8x^3\sin \left(2 x^4+8\right)$. ## Step 4: Combine the derivatives of the two components The derivative of the given function is the sum of the derivatives of its components: $20x^4e^{4 x^5-4} - 8x^3\sin \left(2 x^4+8\right)$. The final answer is: $\boxed{20x^4e^{4 x^5-4} - 8x^3\sin \left(2 x^4+8\right)}$ |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| Qwen2-72B-Instruct | To differentiate the given function, we will apply the chain rule to both terms. Let's start with the first term $e^{4x^5 - 4}$: The derivative of $e^u$ with respect to $u$ is $e^u$, and the derivative of $u = 4x^5 - 4$ with respect to $x$ is $20x^4$. Therefore, by the chain rule, the derivative of $e^{4x^5 - 4}$ with respect to $x$ is $e^{4x^5 - 4} \cdot 20x^4 = 20x^4 e^{4x^5 - 4}$. Next, let's differentiate the second term $\cos(2x^4 + 8)$: The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$, and the derivative of $u = 2x^4 + 8$ with respect to $x$ is $8x^3$. Therefore, by the chain rule, the derivative of $\cos(2x^4 + 8)$ with respect to $x$ is $-\sin(2x^4 + 8) \cdot 8x^3 = -8x^3 \sin(2x^4 + 8)$. Putting it all together, the derivative of the entire function is: $$20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8)$$ So the final answer is: $$\boxed{20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8)}$$ |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| baichuan4 | To differentiate the function $f(x) = e^{4x^5 - 4} + \cos(2x^4 + 8)$, we need to use the chain rule for each term. The chain rule states that if $g(x) = h(u(x))$, then $g'(x) = h'(u(x)) \cdot u'(x)$. For the first term, $e^{4x^5 - 4}$, we have: - $h(u) = e^u$ - $u(x) = 4x^5 - 4$ The derivative of $h(u)$ with respect to $u$ is $h'(u) = e^u$, and the derivative of $u(x)$ with respect to $x$ is $u'(x) = 20x^4$. Applying the chain rule, we get: $$\frac{d}{dx} e^{4x^5 - 4} = e^{4x^5 - 4} \cdot 20x^4$$ For the second term, $\cos(2x^4 + 8)$, we have: - $h(u) = \cos(u)$ - $u(x) = 2x^4 + 8$ The derivative of $h(u)$ with respect to $u$ is $h'(u) = -\sin(u)$, and the derivative of $u(x)$ with respect to $x$ is $u'(x) = 8x^3$. Applying the chain rule, we get: $$\frac{d}{dx} \cos(2x^4 + 8) = -\sin(2x^4 + 8) \cdot 8x^3$$ Now, we combine the derivatives of both terms to get the derivative of $f(x)$: $$f'(x) = e^{4x^5 - 4} \cdot 20x^4 - \sin(2x^4 + 8) \cdot 8x^3$$ Simplifying, we have: $$f'(x) = 20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8)$$ This is the final answer, which we can put in a box as requested: $$\boxed{f'(x) = 20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8)}$$ |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| doubao | Let $f(x)=e^{4x^5-4}+\cos(2x^4+8)$. The derivative of $e^{u}$ is $e^{u}u'$, so for the first term with $u=4x^5-4$, the derivative is $e^{4x^5-4}\times(4\times5x^4)=20x^4e^{4x^5-4}$. The derivative of $\cos(u)$ is $-\sin(u)u'$, so for the second term with $u=2x^4+8$, the derivative is $-\sin(2x^4+8)\times(2\times4x^3)=-8x^3\sin(2x^4+8)$. So the derivative of $f(x)$ is $20x^4e^{4x^5-4}-8x^3\sin(2x^4+8)$. $\boxed{20x^4e^{4x^5-4}-8x^3\sin(2x^4+8)}$. |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| gpto1 | To find the derivative of the function: \[ f(x) = e^{4 x^5 - 4} + \cos(2 x^4 + 8), \] we'll differentiate each term separately. **First term:** \( e^{4 x^5 - 4} \) Let \( u = 4 x^5 - 4 \), so the derivative is: \[ \frac{d}{dx} e^{u} = e^{u} \cdot \frac{du}{dx} = e^{4 x^5 - 4} \cdot (20 x^4). \] **Second term:** \( \cos(2 x^4 + 8) \) Let \( v = 2 x^4 + 8 \), so the derivative is: \[ \frac{d}{dx} \cos(v) = -\sin(v) \cdot \frac{dv}{dx} = -\sin(2 x^4 + 8) \cdot (8 x^3). \] **Combining the derivatives:** \[ f'(x) = 20 x^4 e^{4 x^5 - 4} - 8 x^3 \sin(2 x^4 + 8). \] **Final Answer:** \[\boxed{f'(x)=20x^{4}e^{4x^{5}-4}\,-\,8x^{3}\sin(2x^{4}+8)}\] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| kimi | To differentiate the function \( f(x) = e^{4 x^5 - 4} + \cos(2 x^4 + 8) \), we will use the chain rule for each term separately. ### Step 1: Differentiate \( e^{4 x^5 - 4} \) Let \( u = 4 x^5 - 4 \). Then, \( f(x) = e^u \). The chain rule states that: \[ \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx} \] First, find \( \frac{du}{dx} \): \[ u = 4 x^5 - 4 \] \[ \frac{du}{dx} = 4 \cdot 5 x^4 = 20 x^4 \] Now, apply the chain rule: \[ \frac{d}{dx} e^{4 x^5 - 4} = e^{4 x^5 - 4} \cdot 20 x^4 = 20 x^4 e^{4 x^5 - 4} \] ### Step 2: Differentiate \( \cos(2 x^4 + 8) \) Let \( v = 2 x^4 + 8 \). Then, \( f(x) = \cos(v) \). The chain rule states that: \[ \frac{d}{dx} \cos(v) = -\sin(v) \cdot \frac{dv}{dx} \] First, find \( \frac{dv}{dx} \): \[ v = 2 x^4 + 8 \] \[ \frac{dv}{dx} = 2 \cdot 4 x^3 = 8 x^3 \] Now, apply the chain rule: \[ \frac{d}{dx} \cos(2 x^4 + 8) = -\sin(2 x^4 + 8) \cdot 8 x^3 = -8 x^3 \sin(2 x^4 + 8) \] ### Step 3: Combine the derivatives Add the derivatives of both terms: \[ \frac{d}{dx} \left( e^{4 x^5 - 4} + \cos(2 x^4 + 8) \right) = 20 x^4 e^{4 x^5 - 4} - 8 x^3 \sin(2 x^4 + 8) \] ### Final Answer \[ \boxed{20 x^4 e^{4 x^5 - 4} - 8 x^3 \sin(2 x^4 + 8)} \] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| minimax | To differentiate the function $f(x) = e^{4x^5-4} + \cos(2x^4+8)$, we will apply the chain rule and the derivative of the cosine function. The chain rule states that if you have a composite function $g(h(x))$, then its derivative is $g'(h(x)) \cdot h'(x)$. Let's differentiate the function term by term: 1. For the first term $e^{4x^5-4}$, we treat $4x^5-4$ as the inner function $h(x)$, and $e^u$ as the outer function $g(u)$ where $u = h(x)$. The derivative of $e^u$ with respect to $u$ is $e^u$, and the derivative of $4x^5-4$ with respect to $x$ is $20x^4$. Applying the chain rule, we get: $$ \frac{d}{dx} e^{4x^5-4} = e^{4x^5-4} \cdot \frac{d}{dx}(4x^5-4) = e^{4x^5-4} \cdot 2 |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| wenxinyiyan | To differentiate the given function $e^{4 x^5-4}+\cos \left(2 x^4+8\right)$, we need to apply the chain rule and the basic derivative rules. 1. **Derivative of $e^{4 x^5-4}$**: * The derivative of $e^u$ with respect to $u$ is $e^u$. * Here, $u = 4x^5 - 4$, so $\frac{du}{dx} = 4 \times 5x^4 = 20x^4$. * Applying the chain rule, $\frac{d}{dx}(e^{4 x^5-4}) = e^{4 x^5-4} \times 20x^4$. 2. **Derivative of $\cos \left(2 x^4+8\right)$**: * The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$. * Here, $u = 2x^4 + 8$, so $\frac{du}{dx} = 2 \times 4x^3 = 8x^3$. * Applying the chain rule, $\frac{d}{dx}(\cos(2 x^4+8)) = -\sin(2 x^4+8) \times 8x^3$. Combining these two derivatives, we get: $\frac{d}{dx}(e^{4 x^5-4}+\cos \left(2 x^4+8\right)) = e^{4 x^5-4} \times 20x^4 - \sin(2 x^4+8) \times 8x^3$Finally, we box the answer: $\boxed{e^{4 x^5-4} \times 20x^4 - \sin(2 x^4+8) \times 8x^3}$ |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| zhipu | To differentiate the function \( f(x) = e^{4x^5 - 4} + \cos(2x^4 + 8) \), we will use the chain rule. ### Step-by-Step Differentiation: 1. **Differentiate \( e^{4x^5 - 4} \):** - Let \( u = 4x^5 - 4 \). Then the function becomes \( e^u \). - The derivative of \( e^u \) with respect to \( u \) is \( e^u \). - Now, we need to differentiate \( u = 4x^5 - 4 \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx}(4x^5 - 4) = 20x^4 \] - Using the chain rule, the derivative of \( e^{4x^5 - 4} \) with respect to \( x \) is: \[ \frac{d}{dx} e^{4x^5 - 4} = e^{4x^5 - 4} \cdot \frac{du}{dx} = e^{4x^5 - 4} \cdot 20x^4 \] 2. **Differentiate \( \cos(2x^4 + 8) \):** - Let \( v = 2x^4 + 8 \). Then the function becomes \( \cos(v) \). - The derivative of \( \cos(v) \) with respect to \( v \) is \( -\sin(v) \). - Now, we need to differentiate \( v = 2x^4 + 8 \) with respect to \( x \): \[ \frac{dv}{dx} = \frac{d}{dx}(2x^4 + 8) = 8x^3 \] - Using the chain rule, the derivative of \( \cos(2x^4 + 8) \) with respect to \( x \) is: \[ \frac{d}{dx} \cos(2x^4 + 8) = -\sin(2x^4 + 8) \cdot \frac{dv}{dx} = -\sin(2x^4 + 8) \cdot 8x^3 \] 3. **Combine the results:** - Adding the derivatives of the two parts, we get: \[ f'(x) = \frac{d}{dx} \left( e^{4x^5 - 4} \right) + \frac{d}{dx} \left( \cos(2x^4 + 8) \right) \] \[ f'(x) = e^{4x^5 - 4} \cdot 20x^4 + (-\sin(2x^4 + 8) \cdot 8x^3) \] \[ f'(x) = 20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8) \] ### Final Answer: \[ \boxed{20x^4 e^{4x^5 - 4} - 8x^3 \sin(2x^4 + 8)} \] |
4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-08bb1850-3e8e-9b0a-9083-350e677c6f99', 'request_id': '08bb1850-3e8e-9b0a-9083-350e677c6f99'} | 4 x^3 \left(5 e^{4 x^5-4} x-2 \sin \left(2 \left(x^4+4\right)\right)\right) |