| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the given function \(\sin \left(\frac{13}{2}-8 x^5\right)+e^{\frac{13}{2}-8 x}\), we will use the chain rule. Let's break it down step by step. First, let's denote the function as: \[ f(x) = \sin \left(\frac{13}{2}-8 x^5\right) + e^{\frac{13}{2}-8 x} \] We need to find \( f'(x) \). ### Differentiating the first part: \(\sin \left(\frac{13}{2}-8 x^5\right)\) 1. Let \( u = \frac{13}{2} - 8 x^5 \). 2. Then, \(\sin \left(\frac{13}{2}-8 x^5\right) = \sin(u) \). 3. The derivative of \(\sin(u)\) with respect to \( u \) is \(\cos(u)\). 4. Now, we need to find the derivative of \( u \) with respect to \( x \): \[ \frac{d}{dx} \left( \frac{13}{2} - 8 x^5 \right) = -40 x^4 \] 5. Using the chain rule, the derivative of \(\sin \left(\frac{13}{2}-8 x^5\right)\) with respect to \( x \) is: \[ \frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{du}{dx} = \cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40 x^4) \] ### Differentiating the second part: \( e^{\frac{13}{2}-8 x} \) 1. Let \( v = \frac{13}{2} - 8 x \). 2. Then, \( e^{\frac{13}{2}-8 x} = e^v \). 3. The derivative of \( e^v \) with respect to \( v \) is \( e^v \). 4. Now, we need to find the derivative of \( v \) with respect to \( x \): \[ \frac{d}{dx} \left( \frac{13}{2} - 8 x \right) = -8 \] 5. Using the chain rule, the derivative of \( e^{\frac{13}{2}-8 x} \) with respect to \( x \) is: \[ \frac{d}{dx} e^v = e^v \cdot \frac{dv}{dx} = e^{\frac{13}{2}-8 x} \cdot (-8) \] ### Combining the results Now, we combine the derivatives of the two parts: \[ f'(x) = \cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40 x^4) + e^{\frac{13}{2}-8 x} \cdot (-8) \] Simplifying, we get: \[ f'(x) = -40 x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8 e^{\frac{13}{2}-8 x} \] Thus, the final answer is: \[ \boxed{-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8 e^{\frac{13}{2}-8 x}} \] |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| deepseek-r1 | To differentiate the function \(\sin \left(\frac{13}{2} - 8x^5\right) + e^{\frac{13}{2} - 8x}\), we will use the chain rule for both the sine and exponential terms. 1. **Differentiating the sine term \(\sin \left(\frac{13}{2} - 8x^5\right)\):** - Let \(u = \frac{13}{2} - 8x^5\). - The derivative of \(\sin(u)\) is \(\cos(u) \cdot u'\). - Compute \(u'\): \[ u' = \frac{d}{dx}\left(\frac{13}{2} - 8x^5\right) = 0 - 40x^4 = -40x^4 \] - Therefore, the derivative of the sine term is: \[ \cos\left(\frac{13}{2} - 8x^5\right) \cdot (-40x^4) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) \] 2. **Differentiating the exponential term \(e^{\frac{13}{2} - 8x}\):** - Let \(v = \frac{13}{2} - 8x\). - The derivative of \(e^v\) is \(e^v \cdot v'\). - Compute \(v'\): \[ v' = \frac{d}{dx}\left(\frac{13}{2} - 8x\right) = 0 - 8 = -8 \] - Therefore, the derivative of the exponential term is: \[ e^{\frac{13}{2} - 8x} \cdot (-8) = -8e^{\frac{13}{2} - 8x} \] 3. **Combining the derivatives:** - The derivative of the entire function is the sum of the derivatives of each term: \[ -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x} \] Thus, the derivative of the function is \(\boxed{-40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x}}\). |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| gpto1 | To differentiate the given function \(f(x) = \sin\left(\frac{13}{2} - 8x^5\right) + e^{\frac{13}{2} - 8x}\), we apply the chain rule to each term: **First Term:** Let \(u = \frac{13}{2} - 8x^5\). Then, \[ \frac{d}{dx} \sin(u) = \cos(u) \cdot u' \] Compute \(u'\): \[ u' = \frac{d}{dx} \left(\frac{13}{2} - 8x^5\right) = -40x^4 \] So, \[ \frac{d}{dx} \sin\left(\frac{13}{2} - 8x^5\right) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) \] **Second Term:** Let \(v = \frac{13}{2} - 8x\). Then, \[ \frac{d}{dx} e^{v} = e^{v} \cdot v' \] Compute \(v'\): \[ v' = \frac{d}{dx} \left(\frac{13}{2} - 8x\right) = -8 \] So, \[ \frac{d}{dx} e^{\frac{13}{2} - 8x} = -8 e^{\frac{13}{2} - 8x} \] **Putting it all together:** \[ f'(x) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8 e^{\frac{13}{2} - 8x} \] \(\boxed{\displaystyle f'(x)=-40x^{4}\cos\left(\tfrac{13}{2}-8x^{5}\right)-8e^{\tfrac{13}{2}-8x}}\) |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| gpt-4o-2024-08-06 | To differentiate the function \( f(x) = \sin \left(\frac{13}{2} - 8x^5\right) + e^{\frac{13}{2} - 8x} \), we will apply the chain rule to each term separately. 1. **Differentiate the first term**: \(\sin \left(\frac{13}{2} - 8x^5\right)\). Let \( u = \frac{13}{2} - 8x^5 \). Then, the derivative of \(\sin(u)\) with respect to \( u \) is \(\cos(u)\), and the derivative of \( u \) with respect to \( x \) is: \[ \frac{d}{dx}\left(\frac{13}{2} - 8x^5\right) = -40x^4. \] By the chain rule, the derivative of \(\sin(u)\) with respect to \( x \) is: \[ \cos\left(\frac{13}{2} - 8x^5\right) \cdot (-40x^4) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right). \] 2. **Differentiate the second term**: \(e^{\frac{13}{2} - 8x}\). Let \( v = \frac{13}{2} - 8x \). Then, the derivative of \( e^v \) with respect to \( v \) is \( e^v \), and the derivative of \( v \) with respect to \( x \) is: \[ \frac{d}{dx}\left(\frac{13}{2} - 8x\right) = -8. \] By the chain rule, the derivative of \( e^v \) with respect to \( x \) is: \[ e^{\frac{13}{2} - 8x} \cdot (-8) = -8e^{\frac{13}{2} - 8x}. \] Combining these results, the derivative of the entire function is: \[ f'(x) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x}. \] Thus, the final answer is: \[ \boxed{-40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x}} \] |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| DeepSeek-Coder-V2-Lite-Instruct | The differentiation of the function \( \sin \left(\frac{13}{2}-8 x^5\right) + e^{\frac{13}{2}-8 x} \) involves applying the chain rule to each term separately. 1. **Differentiate \( \sin \left(\frac{13}{2}-8 x^5\right) \):** - The outer function is \( \sin(u) \) where \( u = \frac{13}{2} - 8 x^5 \). - The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). - The derivative of \( u = \frac{13}{2} - 8 x^5 \) with respect to \( x \) is \( -40 x^4 \). - Therefore, the derivative of \( \sin \left(\frac{13}{2}-8 x^5\right) \) with respect to \( x \) is: \[ \cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40 x^4) = -40 x^4 \cos \left(\frac{13}{2}-8 x^5\right) \] 2. **Differentiate \( e^{\frac{13}{2}-8 x} \):** - The outer function is \( e^v \) where \( v = \frac{13}{2} - 8 x \). - The derivative of \( e^v \) with respect to \( v \) is \( e^v \). - The derivative of \( v = \frac{13}{2} - 8 x \) with respect to \( x \) is \( -8 \). - Therefore, the derivative of \( e^{\frac{13}{2}-8 x} \) with respect to \( x \) is: \[ e^{\frac{13}{2}-8 x} \cdot (-8) = -8 e^{\frac{13}{2}-8 x} \] Combining these results, the derivative of the given function is: \[ -40 x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8 e^{\frac{13}{2}-8 x} \] Thus, the final answer is: \[ \boxed{-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8 e^{\frac{13}{2}-8 x}} \] |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| 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: $\sin \left(\frac{13}{2}-8 x^5\right)$ and $e^{\frac{13}{2}-8 x}$. Each of these components needs to be differentiated separately. ## Step 2: Differentiate the sine component using the chain rule To differentiate $\sin \left(\frac{13}{2}-8 x^5\right)$, we use the chain rule. The derivative of $\sin(u)$ is $\cos(u)$, and the derivative of $\frac{13}{2}-8 x^5$ is $-40x^4$. Therefore, the derivative of the sine component is $\cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4)$. ## Step 3: Differentiate the exponential component using the chain rule To differentiate $e^{\frac{13}{2}-8 x}$, we also use the chain rule. The derivative of $e^u$ is $e^u$, and the derivative of $\frac{13}{2}-8 x$ is $-8$. Therefore, the derivative of the exponential component is $e^{\frac{13}{2}-8 x} \cdot (-8)$. ## Step 4: Combine the derivatives of the two components The derivative of the entire function is the sum of the derivatives of its components: $\cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4) + e^{\frac{13}{2}-8 x} \cdot (-8)$. ## Step 5: Simplify the expression if necessary The expression $\cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4) + e^{\frac{13}{2}-8 x} \cdot (-8)$ can be simplified to $-40x^4\cos \left(\frac{13}{2}-8 x^5\right) - 8e^{\frac{13}{2}-8 x}$. The final answer is: $\boxed{-40x^4\cos \left(\frac{13}{2}-8 x^5\right) - 8e^{\frac{13}{2}-8 x}}$ |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the components of the function that need to be differentiated. The function given is a sum of two functions: $\sin \left(\frac{13}{2}-8 x^5\right)$ and $e^{\frac{13}{2}-8 x}$. We will differentiate each component separately. ## Step 2: Differentiate the sine component using the chain rule. To differentiate $\sin \left(\frac{13}{2}-8 x^5\right)$, we use the chain rule. The derivative of $\sin(u)$ is $\cos(u)$, and here $u = \frac{13}{2}-8 x^5$. So, the derivative of the sine component is $\cos \left(\frac{13}{2}-8 x^5\right) \cdot \frac{d}{dx} \left(\frac{13}{2}-8 x^5\right)$. ## Step 3: Calculate the derivative of the inner function of the sine component. The derivative of $\frac{13}{2}-8 x^5$ with respect to $x$ is $-40x^4$. ## Step 4: Combine the results of step 2 and step 3 to get the derivative of the sine component. The derivative of the sine component is $\cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4) = -40x^4 \cos \left(\frac{13}{2}-8 x^5\right)$. ## Step 5: Differentiate the exponential component using the chain rule. To differentiate $e^{\frac{13}{2}-8 x}$, we use the chain rule. The derivative of $e^u$ is $e^u$, and here $u = \frac{13}{2}-8 x$. So, the derivative of the exponential component is $e^{\frac{13}{2}-8 x} \cdot \frac{d}{dx} \left(\frac{13}{2}-8 x\right)$. ## Step 6: Calculate the derivative of the inner function of the exponential component. The derivative of $\frac{13}{2}-8 x$ with respect to $x$ is $-8$. ## Step 7: Combine the results of step 5 and step 6 to get the derivative of the exponential component. The derivative of the exponential component is $e^{\frac{13}{2}-8 x} \cdot (-8) = -8e^{\frac{13}{2}-8 x}$. ## Step 8: Combine the derivatives of the sine and exponential components to get the derivative of the entire function. The derivative of the entire function is the sum of the derivatives of its components: $-40x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8e^{\frac{13}{2}-8 x}$. The final answer is: $\boxed{-40x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8e^{\frac{13}{2}-8 x}}$ |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| Qwen2-72B-Instruct | To differentiate the given function, we will apply the chain rule and the rules for differentiating sine functions and exponential functions. Let's start with the first term: $$\frac{d}{dx}\left[\sin \left(\frac{13}{2}-8 x^5\right)\right]$$ The derivative of $\sin(u)$ is $\cos(u) \cdot u'$, where $u' = \frac{du}{dx}$. Here, $u = \frac{13}{2}-8x^5$, so $u' = -40x^4$. Thus, $$\frac{d}{dx}\left[\sin \left(\frac{13}{2}-8 x^5\right)\right] = \cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4)$$ Now let's differentiate the second term: $$\frac{d}{dx}\left[e^{\frac{13}{2}-8 x}\right]$$ The derivative of $e^u$ is $e^u \cdot u'$, where again $u' = \frac{du}{dx}$. Here, $u = \frac{13}{2}-8x$, so $u' = -8$. Thus, $$\frac{d}{dx}\left[e^{\frac{13}{2}-8 x}\right] = e^{\frac{13}{2}-8 x} \cdot (-8)$$ Putting it all together, the derivative of the entire function is: $$\frac{d}{dx}\left[\sin \left(\frac{13}{2}-8 x^5\right)+e^{\frac{13}{2}-8 x}\right] = \cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4) + e^{\frac{13}{2}-8 x} \cdot (-8)$$ In LaTeX format, this would be written as: $$\boxed{-40x^4\cos \left(\frac{13}{2}-8 x^5\right)-8e^{\frac{13}{2}-8 x}}$$ |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| baichuan4 | To differentiate the given function, we will use the chain rule and the sum rule. The function is composed of two parts: a sine function and an exponential function, both with composite arguments. The function to be differentiated is: $$ f(x) = \sin \left(\frac{13}{2}-8 x^5\right)+e^{\frac{13}{2}-8 x} $$ We will differentiate each part separately. For the sine part, we use the chain rule. Let $u(x) = \frac{13}{2}-8 x^5$. Then, $\sin(u(x))$ has a derivative of $\cos(u(x)) \cdot u'(x)$. To find $u'(x)$, we differentiate $u(x)$ with respect to $x$: $$ u'(x) = -40x^4 $$ Therefore, the derivative of the sine part is: $$ \frac{d}{dx} \sin \left(\frac{13}{2}-8 x^5\right) = \cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4) $$ For the exponential part, we also use the chain rule. Let $v(x) = \frac{13}{2}-8 x$. Then, $e^{v(x)}$ has a derivative of $e^{v(x)} \cdot v'(x)$. To find $v'(x)$, we differentiate $v(x)$ with respect to $x$: $$ v'(x) = -8 $$ Therefore, the derivative of the exponential part is: $$ \frac{d}{dx} e^{\frac{13}{2}-8 x} = e^{\frac{13}{2}-8 x} \cdot (-8) $$ Now, we combine the derivatives of both parts using the sum rule: $$ f'(x) = \cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4) + e^{\frac{13}{2}-8 x} \cdot (-8) $$ Simplifying, we get: $$ f'(x) = -40x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8e^{\frac{13}{2}-8 x} $$ This is the final derivative of the function. We can put it in a box as requested: $$ \boxed{f'(x) = -40x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8e^{\frac{13}{2}-8 x}} $$ |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| doubao | Let $u=\frac{13}{2}-8x^5$ and $v=\frac{13}{2}-8x$. The derivative of $\sin u$ is $\cos u$ times the derivative of $u$: $(\sin u)'=\cos u \cdot u'$. $u'=-40x^4$. So the derivative of $\sin(\frac{13}{2}-8x^5)$ is $\cos(\frac{13}{2}-8x^5)\cdot (-40x^4)=-40x^4\cos(\frac{13}{2}-8x^5)$. The derivative of $e^v$ is $e^v$ times the derivative of $v$: $(e^v)'=e^v \cdot v'$. $v'=-8$. So the derivative of $e^{\frac{13}{2}-8x}$ is $e^{\frac{13}{2}-8x}\cdot (-8)=-8e^{\frac{13}{2}-8x}$. The total derivative is $-40x^4\cos(\frac{13}{2}-8x^5)-8e^{\frac{13}{2}-8x}$. $\boxed{-40x^4\cos(\frac{13}{2}-8x^5)-8e^{\frac{13}{2}-8x}}$. |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| gpto1 | To differentiate the given function \(f(x) = \sin\left(\frac{13}{2} - 8x^5\right) + e^{\frac{13}{2} - 8x}\), we apply the chain rule to each term: **First Term:** Let \(u = \frac{13}{2} - 8x^5\). Then, \[ \frac{d}{dx} \sin(u) = \cos(u) \cdot u' \] Compute \(u'\): \[ u' = \frac{d}{dx} \left(\frac{13}{2} - 8x^5\right) = -40x^4 \] So, \[ \frac{d}{dx} \sin\left(\frac{13}{2} - 8x^5\right) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) \] **Second Term:** Let \(v = \frac{13}{2} - 8x\). Then, \[ \frac{d}{dx} e^{v} = e^{v} \cdot v' \] Compute \(v'\): \[ v' = \frac{d}{dx} \left(\frac{13}{2} - 8x\right) = -8 \] So, \[ \frac{d}{dx} e^{\frac{13}{2} - 8x} = -8 e^{\frac{13}{2} - 8x} \] **Putting it all together:** \[ f'(x) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8 e^{\frac{13}{2} - 8x} \] \(\boxed{\displaystyle f'(x)=-40x^{4}\cos\left(\tfrac{13}{2}-8x^{5}\right)-8e^{\tfrac{13}{2}-8x}}\) |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| kimi | To differentiate the function \( f(x) = \sin \left(\frac{13}{2} - 8x^5\right) + e^{\frac{13}{2} - 8x} \), we will use the chain rule for each term separately. ### Step 1: Differentiate \(\sin \left(\frac{13}{2} - 8x^5\right)\) Let \( u = \frac{13}{2} - 8x^5 \). Then, \( f(x) = \sin(u) \). Using the chain rule: \[ \frac{d}{dx}[\sin(u)] = \cos(u) \cdot \frac{du}{dx} \] First, find \(\frac{du}{dx}\): \[ u = \frac{13}{2} - 8x^5 \] \[ \frac{du}{dx} = \frac{d}{dx}\left(\frac{13}{2}\right) - \frac{d}{dx}(8x^5) = 0 - 40x^4 = -40x^4 \] Now, substitute back: \[ \frac{d}{dx}[\sin(u)] = \cos\left(\frac{13}{2} - 8x^5\right) \cdot (-40x^4) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) \] ### Step 2: Differentiate \(e^{\frac{13}{2} - 8x}\) Let \( v = \frac{13}{2} - 8x \). Then, \( f(x) = e^v \). Using the chain rule: \[ \frac{d}{dx}[e^v] = e^v \cdot \frac{dv}{dx} \] First, find \(\frac{dv}{dx}\): \[ v = \frac{13}{2} - 8x \] \[ \frac{dv}{dx} = \frac{d}{dx}\left(\frac{13}{2}\right) - \frac{d}{dx}(8x) = 0 - 8 = -8 \] Now, substitute back: \[ \frac{d}{dx}[e^v] = e^{\frac{13}{2} - 8x} \cdot (-8) = -8e^{\frac{13}{2} - 8x} \] ### Step 3: Combine the derivatives Add the derivatives from both parts: \[ f'(x) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x} \] Thus, the final answer is: \[ \boxed{-40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x}} \] |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| minimax | To differentiate the given function $f(x) = \sin \left(\frac{13}{2}-8 x^5\right)+e^{\frac{13}{2}-8 x}$, we will apply the chain rule and the exponential function differentiation rule. The chain rule states that if you have a composite function $g(h(x))$, then its derivative is $g'(h(x)) \cdot h'(x)$. The derivative of the sine function is the cosine function, and the derivative of the exponential function $e^u$ with respect to $u$ is just $e^u$. Let's differentiate each term separately: 1. For the first term $\sin \left(\frac{13}{2}-8 x^5\right)$, let $u = \frac{13}{2}-8 x^5$. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$, and the derivative of $u$ with respect to $x$ is $-40 x^4$. So, the derivative of the first term with respect to $x$ is: $$ \frac{d}{dx} \sin \left(\frac{13}{2}-8 x |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| wenxinyiyan | The derivative of the function $\sin \left(\frac{13}{2}-8 x^5\right)+e^{\frac{13}{2}-8 x}$ is found using the chain rule and the derivatives of the sine and exponential functions. For the first term, $\sin \left(\frac{13}{2}-8 x^5\right)$, we apply the chain rule with the outer function being sine and the inner function being $\frac{13}{2}-8 x^5$. The derivative of sine is cosine, and the derivative of the inner function is $-40x^4$ (using the power rule). So, the derivative of the first term is $\cos \left(\frac{13}{2}-8 x^5\right) \cdot (-40x^4)$. For the second term, $e^{\frac{13}{2}-8 x}$, we again apply the chain rule with the outer function being exponential and the inner function being $\frac{13}{2}-8 x$. The derivative of the exponential function is itself, and the derivative of the inner function is $-8$. So, the derivative of the second term is $e^{\frac{13}{2}-8 x} \cdot (-8)$. Adding these two derivatives together, we get: $$\boxed{\frac{d}{dx}\left(\sin \left(\frac{13}{2}-8 x^5\right)+e^{\frac{13}{2}-8 x}\right) = -40x^4 \cos \left(\frac{13}{2}-8 x^5\right) - 8e^{\frac{13}{2}-8 x}}$$ |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| zhipu | To differentiate the function \( f(x) = \sin \left(\frac{13}{2} - 8x^5\right) + e^{\frac{13}{2} - 8x} \), we will use the chain rule for both terms. 1. **Differentiate the first term \( \sin \left(\frac{13}{2} - 8x^5\right) \):** Let \( u = \frac{13}{2} - 8x^5 \). Then the first term is \( \sin(u) \). The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). Now, we need to find the derivative of \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx} \left(\frac{13}{2} - 8x^5\right) = -40x^4. \] Using the chain rule, the derivative of \( \sin(u) \) with respect to \( x \) is: \[ \frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{du}{dx} = \cos\left(\frac{13}{2} - 8x^5\right) \cdot (-40x^4). \] 2. **Differentiate the second term \( e^{\frac{13}{2} - 8x} \):** Let \( v = \frac{13}{2} - 8x \). Then the second term is \( e^v \). The derivative of \( e^v \) with respect to \( v \) is \( e^v \). Now, we need to find the derivative of \( v \) with respect to \( x \): \[ \frac{dv}{dx} = \frac{d}{dx} \left(\frac{13}{2} - 8x\right) = -8. \] Using the chain rule, the derivative of \( e^v \) with respect to \( x \) is: \[ \frac{d}{dx} e^v = e^v \cdot \frac{dv}{dx} = e^{\frac{13}{2} - 8x} \cdot (-8). \] 3. **Combine the derivatives:** The derivative of the original function \( f(x) \) is the sum of the derivatives of the two terms: \[ f'(x) = \cos\left(\frac{13}{2} - 8x^5\right) \cdot (-40x^4) + e^{\frac{13}{2} - 8x} \cdot (-8). \] Simplifying, we get: \[ f'(x) = -40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x}. \] Therefore, the final answer is: \[ \boxed{-40x^4 \cos\left(\frac{13}{2} - 8x^5\right) - 8e^{\frac{13}{2} - 8x}}. \] |
-40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-ce510ce1-7c30-91b5-acc4-248398134bbd', 'request_id': 'ce510ce1-7c30-91b5-acc4-248398134bbd'} | -40 x^4 \cos \left(\frac{13}{2}-8 x^5\right)-8 e^{\frac{13}{2}-8 x} |