| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the function \(\tan (5 x+1) - \cos (7-8 x)\), we will use the chain rule for differentiation. Let's break it down step by step. 1. **Differentiate \(\tan (5 x+1)\):** - The derivative of \(\tan u\) is \(\sec^2 u\), where \(u = 5x + 1\). - Using the chain rule, the derivative of \(\tan (5 x+1)\) is: \[ \frac{d}{dx} \tan (5 x+1) = \sec^2 (5 x+1) \cdot \frac{d}{dx} (5 x+1) \] - The derivative of \(5x + 1\) with respect to \(x\) is 5. - Therefore, the derivative of \(\tan (5 x+1)\) is: \[ \sec^2 (5 x+1) \cdot 5 = 5 \sec^2 (5 x+1) \] 2. **Differentiate \(-\cos (7-8 x)\):** - The derivative of \(-\cos u\) is \(\sin u\), where \(u = 7 - 8x\). - Using the chain rule, the derivative of \(-\cos (7-8 x)\) is: \[ \frac{d}{dx} (-\cos (7-8 x)) = -\sin (7-8 x) \cdot \frac{d}{dx} (7-8 x) \] - The derivative of \(7 - 8x\) with respect to \(x\) is -8. - Therefore, the derivative of \(-\cos (7-8 x)\) is: \[ -\sin (7-8 x) \cdot (-8) = 8 \sin (7-8 x) \] 3. **Combine the results:** - The derivative of the entire function \(\tan (5 x+1) - \cos (7-8 x)\) is the sum of the derivatives of each part: \[ \frac{d}{dx} (\tan (5 x+1) - \cos (7-8 x)) = 5 \sec^2 (5 x+1) + 8 \sin (7-8 x) \] Thus, the final answer is: \[ \boxed{5 \sec^2 (5 x+1) + 8 \sin (7-8 x)} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| deepseek-r1 | To differentiate the function \( \tan(5x + 1) - \cos(7 - 8x) \), we apply the chain rule to each term separately. 1. **Differentiating \( \tan(5x + 1) \)**: - Let \( u = 5x + 1 \). The derivative of \( \tan(u) \) with respect to \( x \) is \( \sec^2(u) \cdot \frac{du}{dx} \). - \( \frac{du}{dx} = 5 \). - Therefore, the derivative is \( 5 \sec^2(5x + 1) \). 2. **Differentiating \( -\cos(7 - 8x) \)**: - Let \( v = 7 - 8x \). The derivative of \( -\cos(v) \) with respect to \( x \) is \( \sin(v) \cdot \frac{dv}{dx} \). - \( \frac{dv}{dx} = -8 \). - Therefore, the derivative is \( -8 \sin(7 - 8x) \). Combining these results, the derivative of the function \( \tan(5x + 1) - \cos(7 - 8x) \) is: \[ \boxed{5 \sec^2(5x + 1) - 8 \sin(7 - 8x)} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| gpto1 | To find the derivative of the function \( f(x) = \tan (5x + 1) - \cos(7 - 8x) \), we will compute the derivative of each term individually and then combine the results. **First Term: \( \tan (5x + 1) \)** Let’s let \( u = 5x + 1 \). The derivative of \( \tan u \) with respect to \( x \) is \( \sec^2 u \cdot \frac{du}{dx} \). Compute \( \frac{du}{dx} = 5 \). Then, \[ \frac{d}{dx} \tan(5x + 1) = \sec^2(5x + 1) \cdot 5 = 5 \sec^2(5x + 1). \] **Second Term: \( -\cos(7 - 8x) \)** Let’s let \( v = 7 - 8x \). The derivative of \( -\cos v \) with respect to \( x \) is \( \sin v \cdot \frac{dv}{dx} \). Compute \( \frac{dv}{dx} = -8 \). Then, \[ \frac{d}{dx} [-\cos(7 - 8x)] = \sin(7 - 8x) \cdot (-8) = -8 \sin(7 - 8x). \] **Combined Derivative:** Adding the derivatives of the two terms gives the derivative of the entire function: \[ f'(x) = 5 \sec^2(5x + 1) - 8 \sin(7 - 8x). \] **Final Answer:** \[ \boxed{\,5\,\sec^2(5x+1)\ -\ 8\,\sin(7-8x)\,} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| gpt-4o-2024-08-06 | To differentiate the function \( f(x) = \tan(5x + 1) - \cos(7 - 8x) \), we will differentiate each term separately. 1. **Differentiate \(\tan(5x + 1)\):** The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\). Therefore, using the chain rule, the derivative of \(\tan(5x + 1)\) with respect to \(x\) is: \[ \frac{d}{dx}[\tan(5x + 1)] = \sec^2(5x + 1) \cdot \frac{d}{dx}(5x + 1). \] The derivative of \(5x + 1\) with respect to \(x\) is 5. Thus, we have: \[ \frac{d}{dx}[\tan(5x + 1)] = 5 \sec^2(5x + 1). \] 2. **Differentiate \(-\cos(7 - 8x)\):** The derivative of \(-\cos(u)\) with respect to \(u\) is \(\sin(u)\). Therefore, using the chain rule, the derivative of \(-\cos(7 - 8x)\) with respect to \(x\) is: \[ \frac{d}{dx}[-\cos(7 - 8x)] = \sin(7 - 8x) \cdot \frac{d}{dx}(-8x). \] The derivative of \(-8x\) with respect to \(x\) is \(-8\). Thus, we have: \[ \frac{d}{dx}[-\cos(7 - 8x)] = 8 \sin(7 - 8x). \] 3. **Combine the derivatives:** The derivative of the entire function \( f(x) = \tan(5x + 1) - \cos(7 - 8x) \) is: \[ f'(x) = 5 \sec^2(5x + 1) + 8 \sin(7 - 8x). \] Thus, the derivative of the function is: \[ \boxed{5 \sec^2(5x + 1) + 8 \sin(7 - 8x)} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| DeepSeek-Coder-V2-Lite-Instruct | The differentiation of the function \(\tan (5 x+1) - \cos (7-8 x)\) involves differentiating each term separately. 1. Differentiate \(\tan (5 x+1)\): \[ \frac{d}{dx}[\tan (5 x+1)] = \sec^2(5 x+1) \cdot \frac{d}{dx}[5 x+1] = 5 \sec^2(5 x+1) \] 2. Differentiate \(\cos (7-8 x)\): \[ \frac{d}{dx}[\cos (7-8 x)] = -\sin (7-8 x) \cdot \frac{d}{dx}[7-8 x] = -(-8) \sin (7-8 x) = 8 \sin (7-8 x) \] Combining these results, the derivative of the function is: \[ \frac{d}{dx}[\tan (5 x+1) - \cos (7-8 x)] = 5 \sec^2(5 x+1) + 8 \sin (7-8 x) \] Thus, the final answer is: \[ \boxed{5 \sec^2(5 x+1) + 8 \sin (7-8 x)} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Recall the derivatives of basic trigonometric functions The derivative of $\tan(u)$ with respect to $u$ is $\sec^2(u)$, and the derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$. ## Step 2: Apply the chain rule for differentiation To differentiate $\tan(5x+1)$ and $\cos(7-8x)$, we need to apply the chain rule. The chain rule states that if we have a composite function $f(g(x))$, its derivative is $f'(g(x)) \cdot g'(x)$. ## Step 3: Differentiate $\tan(5x+1)$ using the chain rule Let $u = 5x + 1$. Then, $\frac{du}{dx} = 5$. The derivative of $\tan(5x+1)$ is $\sec^2(5x+1) \cdot 5$. ## Step 4: Differentiate $\cos(7-8x)$ using the chain rule Let $v = 7 - 8x$. Then, $\frac{dv}{dx} = -8$. The derivative of $\cos(7-8x)$ is $-\sin(7-8x) \cdot (-8)$. ## Step 5: Combine the results The derivative of the given function is the derivative of $\tan(5x+1)$ minus the derivative of $\cos(7-8x)$. So, the derivative is $5\sec^2(5x+1) - (-8)\sin(7-8x)$. ## Step 6: Simplify the expression Simplify the expression to get the final derivative: $5\sec^2(5x+1) + 8\sin(7-8x)$. The final answer is: $\boxed{5\sec^2(5x+1) + 8\sin(7-8x)}$ |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the functions to be differentiated The given function is $\tan (5 x+1)-\cos (7-8 x)$. We need to differentiate this function with respect to $x$. ## Step 2: Apply the chain rule for differentiation To differentiate $\tan (5 x+1)$, we'll use the chain rule. The derivative of $\tan u$ is $\sec^2 u$, and we need to multiply this by the derivative of $u$, which is $5$. ## Step 3: Differentiate the first term The derivative of $\tan (5 x+1)$ is $5 \sec^2 (5 x+1)$. ## Step 4: Apply the chain rule for the second term To differentiate $\cos (7-8 x)$, we'll use the chain rule. The derivative of $\cos u$ is $-\sin u$, and we need to multiply this by the derivative of $u$, which is $-8$. ## Step 5: Differentiate the second term The derivative of $\cos (7-8 x)$ is $8 \sin (7-8 x)$. ## Step 6: Combine the derivatives of both terms The derivative of the given function is $5 \sec^2 (5 x+1) - 8 \sin (7-8 x)$. ## Step 7: Write the final answer in the required format The final answer is: $\boxed{5 \sec^2 (5 x+1) - 8 \sin (7-8 x)}$ |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| Qwen2-72B-Instruct | To differentiate the given function, we will apply the chain rule to each term separately. The function is: \[f(x) = \tan(5x + 1) - \cos(7 - 8x)\] Let's differentiate term by term: For the first term, $\tan(5x + 1)$: \[\frac{d}{dx} \tan(5x + 1) = \sec^2(5x + 1) \cdot \frac{d}{dx}(5x + 1) = 5\sec^2(5x + 1)\] For the second term, $-\cos(7 - 8x)$: \[\frac{d}{dx} -\cos(7 - 8x) = \sin(7 - 8x) \cdot \frac{d}{dx}(7 - 8x) = 8\sin(7 - 8x)\] So, the derivative of the entire function is: \[f'(x) = 5\sec^2(5x + 1) + 8\sin(7 - 8x)\] Therefore, the final answer is: \[ \boxed{f'(x) = 5\sec^2(5x + 1) + 8\sin(7 - 8x)} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| baichuan4 | To differentiate the function $\tan(5x+1) - \cos(7-8x)$, we need to use the chain rule and the basic derivatives of trigonometric functions. First, we know that the derivative of $\tan(u)$ is $\sec^2(u)$ and the derivative of $\cos(u)$ is $-\sin(u)$. Here, $u$ represents a function of $x$. Applying the chain rule, we get: 1. For $\tan(5x+1)$: - Let $u = 5x + 1$. - Then $\frac{du}{dx} = 5$. - The derivative of $\tan(u)$ is $\sec^2(u)$. - So, the derivative of $\tan(5x+1)$ is $\sec^2(5x+1) \cdot \frac{du}{dx} = 5\sec^2(5x+1)$. 2. For $-\cos(7-8x)$: - Let $v = 7 - 8x$. - Then $\frac{dv}{dx} = -8$. - The derivative of $\cos(v)$ is $-\sin(v)$. - So, the derivative of $-\cos(7-8x)$ is $-(-\sin(7-8x)) \cdot \frac{dv}{dx} = 8\sin(7-8x)$. Putting these two derivatives together, we get the derivative of the original function: $$ \frac{d}{dx}[\tan(5x+1) - \cos(7-8x)] = 5\sec^2(5x+1) + 8\sin(7-8x). $$ Therefore, the final answer is: $$ \boxed{5\sec^2(5x+1) + 8\sin(7-8x)}. $$ |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| doubao | The derivative of $\tan u$ is $\sec^2 u \cdot u'$. Let $u=5x+1$, then the derivative of $\tan(5x+1)$ is $\sec^2(5x+1) \cdot 5$. The derivative of $\cos u$ is $-\sin u \cdot u'$. Let $u=7-8x$, then the derivative of $-\cos(7-8x)$ is $-(-\sin(7-8x)) \cdot (-8) = 8\sin(7-8x)$. So the derivative of the given function is $\boxed{5\sec^2(5x+1) + 8\sin(7-8x)}$. |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| gpto1 | To find the derivative of the function \( f(x) = \tan (5x + 1) - \cos(7 - 8x) \), we will compute the derivative of each term individually and then combine the results. **First Term: \( \tan (5x + 1) \)** Let’s let \( u = 5x + 1 \). The derivative of \( \tan u \) with respect to \( x \) is \( \sec^2 u \cdot \frac{du}{dx} \). Compute \( \frac{du}{dx} = 5 \). Then, \[ \frac{d}{dx} \tan(5x + 1) = \sec^2(5x + 1) \cdot 5 = 5 \sec^2(5x + 1). \] **Second Term: \( -\cos(7 - 8x) \)** Let’s let \( v = 7 - 8x \). The derivative of \( -\cos v \) with respect to \( x \) is \( \sin v \cdot \frac{dv}{dx} \). Compute \( \frac{dv}{dx} = -8 \). Then, \[ \frac{d}{dx} [-\cos(7 - 8x)] = \sin(7 - 8x) \cdot (-8) = -8 \sin(7 - 8x). \] **Combined Derivative:** Adding the derivatives of the two terms gives the derivative of the entire function: \[ f'(x) = 5 \sec^2(5x + 1) - 8 \sin(7 - 8x). \] **Final Answer:** \[ \boxed{\,5\,\sec^2(5x+1)\ -\ 8\,\sin(7-8x)\,} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| kimi | 1. **Identify the function to differentiate:** \[ f(x) = \tan(5x + 1) - \cos(7 - 8x) \] 2. **Differentiate each term separately using the chain rule:** - **Differentiate \(\tan(5x + 1)\):** \[ \frac{d}{dx} \left[ \tan(5x + 1) \right] = \sec^2(5x + 1) \cdot \frac{d}{dx}(5x + 1) \] Since \(\frac{d}{dx}(5x + 1) = 5\), we get: \[ \frac{d}{dx} \left[ \tan(5x + 1) \right] = 5 \sec^2(5x + 1) \] - **Differentiate \(\cos(7 - 8x)\):** \[ \frac{d}{dx} \left[ \cos(7 - 8x) \right] = -\sin(7 - 8x) \cdot \frac{d}{dx}(7 - 8x) \] Since \(\frac{d}{dx}(7 - 8x) = -8\), we get: \[ \frac{d}{dx} \left[ \cos(7 - 8x) \right] = 8 \sin(7 - 8x) \] 3. **Combine the derivatives of the individual terms:** \[ f'(x) = \frac{d}{dx} \left[ \tan(5x + 1) \right] - \frac{d}{dx} \left[ \cos(7 - 8x) \right] \] Substituting the derivatives: \[ f'(x) = 5 \sec^2(5x + 1) + 8 \sin(7 - 8x) \] 4. **Final answer:** \[ \boxed{5 \sec^2(5x + 1) + 8 \sin(7 - 8x)} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| minimax | To differentiate the function $f(x) = \tan (5x+1) - \cos (7-8x)$, we will apply the chain rule and the basic differentiation rules for trigonometric functions. 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 $\tan(u)$ with respect to $u$ is $\sec^2(u)$, and the derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$. Let's differentiate the function term by term: 1. For the first term $\tan (5x+1)$, let $u = 5x+1$. Then the derivative of this term with respect to $x$ is $\sec^2(u) \cdot \frac{du}{dx}$. Since $u = 5x+1$, $\frac{du}{dx} = 5$. So the derivative of the first term is $5\sec^2(5x+1)$. 2. For the second term $-\cos (7-8x)$, let $v = 7-8x$. Then the derivative of this |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| wenxinyiyan | The derivative of $\tan (5 x+1)$ is $5 \sec ^2(5 x+1)$ and the derivative of $\cos (7-8 x)$ is $8 \sin (7-8 x)$. So, the derivative of $\tan (5 x+1)-\cos (7-8 x)$ is $5 \sec ^2(5 x+1)+8 \sin (7-8 x)$. Final answer: $\boxed{5 \sec ^2(5 x+1)+8 \sin (7-8 x)}$ |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| zhipu | To differentiate the function \( f(x) = \tan(5x + 1) - \cos(7 - 8x) \), we will use the chain rule for each term. 1. **Differentiate \(\tan(5x + 1)\):** The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\), where \(u = 5x + 1\). Using the chain rule, we multiply by the derivative of \(u\) with respect to \(x\): \[ \frac{d}{dx} \tan(5x + 1) = \sec^2(5x + 1) \cdot \frac{d}{dx}(5x + 1) = \sec^2(5x + 1) \cdot 5. \] 2. **Differentiate \(\cos(7 - 8x)\):** The derivative of \(\cos(u)\) with respect to \(u\) is \(-\sin(u)\), where \(u = 7 - 8x\). Using the chain rule, we multiply by the derivative of \(u\) with respect to \(x\): \[ \frac{d}{dx} \cos(7 - 8x) = -\sin(7 - 8x) \cdot \frac{d}{dx}(7 - 8x) = -\sin(7 - 8x) \cdot (-8) = 8\sin(7 - 8x). \] 3. **Combine the results:** Putting it all together, we have: \[ f'(x) = \frac{d}{dx} \left( \tan(5x + 1) - \cos(7 - 8x) \right) = 5\sec^2(5x + 1) + 8\sin(7 - 8x). \] Thus, the derivative of the function is: \[ \boxed{5\sec^2(5x + 1) + 8\sin(7 - 8x)}. \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |
| deepseek-v3 | To differentiate the function \( \tan(5x + 1) - \cos(7 - 8x) \), we'll differentiate each term separately using the chain rule. 1. **Differentiate \( \tan(5x + 1) \):** \[ \frac{d}{dx} [\tan(5x + 1)] = \sec^2(5x + 1) \cdot \frac{d}{dx}(5x + 1) \] \[ = \sec^2(5x + 1) \cdot 5 \] \[ = 5\sec^2(5x + 1) \] 2. **Differentiate \( -\cos(7 - 8x) \):** \[ \frac{d}{dx} [-\cos(7 - 8x)] = \sin(7 - 8x) \cdot \frac{d}{dx}(7 - 8x) \] \[ = \sin(7 - 8x) \cdot (-8) \] \[ = -8\sin(7 - 8x) \] 3. **Combine the results:** \[ \frac{d}{dx} [\tan(5x + 1) - \cos(7 - 8x)] = 5\sec^2(5x + 1) - 8\sin(7 - 8x) \] **Final Answer:** \[ \boxed{5\sec^{2}(5x + 1) - 8\sin(7 - 8x)} \] |
5 \sec ^2(5 x+1)-8 \sin (7-8 x) |