| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the given function \( -2x + \tan \left( \frac{9}{2} - \frac{17x}{2} \right) + \frac{3}{2} \), we will follow these steps: 1. **Differentiate each term separately:** - The first term is \(-2x\). - The second term is \(\tan \left( \frac{9}{2} - \frac{17x}{2} \right)\). - The third term is \(\frac{3}{2}\), which is a constant. 2. **Differentiate the first term:** \[ \frac{d}{dx}(-2x) = -2 \] 3. **Differentiate the second term:** \[ \frac{d}{dx} \left( \tan \left( \frac{9}{2} - \frac{17x}{2} \right) \right) \] To differentiate this, we use the chain rule. Let \( u = \frac{9}{2} - \frac{17x}{2} \). Then, \[ \frac{d}{dx} \left( \tan(u) \right) = \sec^2(u) \cdot \frac{du}{dx} \] First, find \(\frac{du}{dx}\): \[ u = \frac{9}{2} - \frac{17x}{2} \implies \frac{du}{dx} = -\frac{17}{2} \] Now, substitute back: \[ \frac{d}{dx} \left( \tan \left( \frac{9}{2} - \frac{17x}{2} \right) \right) = \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right) \cdot \left( -\frac{17}{2} \right) \] 4. **Differentiate the third term:** \[ \frac{d}{dx} \left( \frac{3}{2} \right) = 0 \] 5. **Combine the results:** \[ \frac{d}{dx} \left( -2x + \tan \left( \frac{9}{2} - \frac{17x}{2} \right) + \frac{3}{2} \right) = -2 + \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right) \cdot \left( -\frac{17}{2} \right) \] Simplify the expression: \[ -2 - \frac{17}{2} \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right) \] Thus, the final differentiated function is: \[ \boxed{-2 - \frac{17}{2} \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right)} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| deepseek-r1 | To differentiate the function \(-2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2}\), we proceed term by term: 1. The derivative of \(-2x\) with respect to \(x\) is \(-2\). 2. For the term \(\tan\left(\frac{9}{2} - \frac{17x}{2}\right)\), we use the chain rule. Let \(u = \frac{9}{2} - \frac{17x}{2}\). The derivative of \(\tan(u)\) is \(\sec^2(u) \cdot u'\). The derivative of \(u\) with respect to \(x\) is \(-\frac{17}{2}\). Therefore, the derivative of \(\tan\left(\frac{9}{2} - \frac{17x}{2}\right)\) is \(-\frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)\). 3. The derivative of the constant term \(\frac{3}{2}\) is \(0\). Combining these results, the derivative of the function is: \[ -2 - \frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \] \[ \boxed{-2 - \frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| gpto1 | To find \( f'(x) \) for the function \( f(x) = -2x + \tan\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) + \dfrac{3}{2} \), we'll proceed step by step. **Step 1: Identify the inner function and its derivative.** Let \[ u(x) = \dfrac{9}{2} - \dfrac{17x}{2} \] Then \[ u'(x) = -\dfrac{17}{2} \] **Step 2: Differentiate each term of \( f(x) \) separately.** The derivative of \( -2x \) is \( -2 \). The derivative of \( \dfrac{3}{2} \) is \( 0 \) (since it's a constant). For the term involving the tangent function, use the chain rule: \[ \frac{d}{dx} \left[ \tan(u(x)) \right] = u'(x) \sec^2(u(x)) \] So, \[ \frac{d}{dx} \left[ \tan\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) \right] = -\dfrac{17}{2} \sec^2\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) \] **Step 3: Combine the derivatives.** Adding up the derivatives: \[ f'(x) = -2 - \dfrac{17}{2} \sec^2\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) \] **Final Answer:** \[ \boxed{\, -2\ -\ \dfrac{17}{2}\ \sec^{2}\left(\dfrac{9}{2}-\dfrac{17x}{2}\right) \,} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| gpt-4o-2024-08-06 | To differentiate the function \( f(x) = -2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2} \), we will differentiate each term separately. 1. The derivative of \(-2x\) with respect to \(x\) is \(-2\). 2. The derivative of \(\frac{3}{2}\) with respect to \(x\) is \(0\) since it is a constant. 3. For the term \(\tan\left(\frac{9}{2} - \frac{17x}{2}\right)\), we use the chain rule. The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\). Let \( u = \frac{9}{2} - \frac{17x}{2} \). Then, the derivative of \(u\) with respect to \(x\) is \(-\frac{17}{2}\). Applying the chain rule, the derivative of \(\tan\left(\frac{9}{2} - \frac{17x}{2}\right)\) with respect to \(x\) is: \[ \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right) = -\frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right). \] Combining these results, the derivative of the entire function is: \[ -2 + 0 - \frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right). \] Thus, the derivative of the function is: \[ \boxed{-2 - \frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| DeepSeek-Coder-V2-Lite-Instruct | The differentiation of the function $-2 x+\tan \left(\frac{9}{2}-\frac{17 x}{2}\right)+\frac{3}{2}$ can be found by differentiating each term separately. 1. The derivative of $-2x$ is $-2$. 2. The derivative of $\tan \left(\frac{9}{2}-\frac{17 x}{2}\right)$ requires the chain rule. Let $u = \frac{9}{2} - \frac{17 x}{2}$. Then, the derivative of $\tan(u)$ with respect to $u$ is $\sec^2(u)$. The derivative of $u$ with respect to $x$ is $-\frac{17}{2}$. Therefore, the derivative of $\tan \left(\frac{9}{2}-\frac{17 x}{2}\right)$ is $\sec^2 \left(\frac{9}{2} - \frac{17 x}{2}\right) \cdot \left(-\frac{17}{2}\right) = -\frac{17}{2} \sec^2 \left(\frac{9}{2} - \frac{17 x}{2}\right)$. 3. The derivative of $\frac{3}{2}$ is $0$. Combining these results, the derivative of the function is: \[ -2 - \frac{17}{2} \sec^2 \left(\frac{9}{2} - \frac{17 x}{2}\right) \] Thus, the final answer is: \[ \boxed{-2 - \frac{17}{2} \sec^2 \left(\frac{9}{2} - \frac{17 x}{2}\right)} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the components of the function that need differentiation The given function is $-2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2}$. We need to differentiate each component separately. ## Step 2: Differentiate the linear component The derivative of $-2x$ is $-2$. ## Step 3: Differentiate the tangent component using the chain rule The derivative of $\tan(u)$ is $\sec^2(u) \cdot \frac{du}{dx}$. Here, $u = \frac{9}{2} - \frac{17x}{2}$, so $\frac{du}{dx} = -\frac{17}{2}$. Thus, the derivative of the tangent component is $\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right)$. ## Step 4: Differentiate the constant component The derivative of $\frac{3}{2}$ is $0$. ## Step 5: Combine the derivatives of all components The derivative of the entire function is $-2 + \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right) + 0$. ## Step 6: Simplify the expression The derivative simplifies to $-2 - \frac{17}{2}\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)$. The final answer is: $\boxed{-2 - \frac{17}{2}\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)}$ |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the components of the function that need to be differentiated. The function given is $-2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2}$. This function has three components: a linear term $-2x$, a trigonometric term $\tan\left(\frac{9}{2} - \frac{17x}{2}\right)$, and a constant term $\frac{3}{2}$. ## Step 2: Differentiate the linear term. The derivative of $-2x$ with respect to $x$ is $-2$. ## Step 3: Differentiate the trigonometric term using the chain rule. To differentiate $\tan\left(\frac{9}{2} - \frac{17x}{2}\right)$, we use the chain rule. The derivative of $\tan(u)$ is $\sec^2(u)$, and here $u = \frac{9}{2} - \frac{17x}{2}$. The derivative of $u$ with respect to $x$ is $-\frac{17}{2}$. Therefore, the derivative of the trigonometric term is $\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right)$. ## Step 4: Differentiate the constant term. The derivative of a constant is $0$, so the derivative of $\frac{3}{2}$ is $0$. ## Step 5: Combine the derivatives of all components to get the final derivative. Combining the derivatives from the previous steps, we get $-2 + \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right) + 0$. ## Step 6: Simplify the expression if necessary. The expression $-2 + \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right)$ is already simplified in terms of the given function's components. The final answer is: $\boxed{-2 -\frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)}$ |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| Qwen2-72B-Instruct | To differentiate the given function, we will apply basic rules of differentiation to each term separately. The function is: \[f(x) = -2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2}\] Let's differentiate term by term: 1. The derivative of $-2x$ with respect to $x$ is $-2$. 2. For the $\tan\left(\frac{9}{2} - \frac{17x}{2}\right)$ part, we use the chain rule. The derivative of $\tan(u)$ is $\sec^2(u)$ and the derivative of $u=\frac{9}{2} - \frac{17x}{2}$ is $-\frac{17}{2}$. So, the derivative of this term is $\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right)$. 3. The derivative of the constant $\frac{3}{2}$ is $0$ since the derivative of any constant is zero. Putting it all together, the derivative of the function is: \[f'(x) = -2 + \left(-\frac{17}{2}\right)\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)\] So, the final answer is: \[f'(x) = -2 - \frac{17}{2}\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)\] In $\boxed{}$ format: \[\boxed{-2 - \frac{17}{2}\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)}\] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| baichuan4 | To differentiate the given function, we need to use the basic differentiation rules, including the power rule, the constant rule, and the chain rule for trigonometric functions. The function to be differentiated is: $$ f(x) = -2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2} $$ We will differentiate each term separately. 1. For the first term $-2x$, the derivative is simply $-2$. 2. For the second term $\tan\left(\frac{9}{2} - \frac{17x}{2}\right)$, we need to use the chain rule. The derivative of $\tan(u)$ with respect to $u$ is $\sec^2(u)$, where $u = \frac{9}{2} - \frac{17x}{2}$. Now we need to find the derivative of $u$ with respect to $x$, which is $-\frac{17}{2}$. Applying the chain rule, we get: $$ \frac{d}{dx} \tan\left(\frac{9}{2} - \frac{17x}{2}\right) = \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right) $$ 3. For the third term $\frac{3}{2}$, which is a constant, the derivative is $0$. Now, we combine these derivatives to get the final derivative of the function: $$ f'(x) = -2 + \left(-\frac{17}{2}\right) \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) $$ Simplifying the expression, we get: $$ f'(x) = -2 - \frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) $$ This can also be written as: $$ f'(x) = -2 - \frac{17}{2} \sec^2\left(\frac{1}{2}(9 - 17x)\right) $$ And finally, we can put the answer in a box as requested: $$ \boxed{f'(x) = -2 - \frac{17}{2} \sec^2\left(\frac{1}{2}(9 - 17x)\right)} $$ |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| doubao | Let $f(x)=-2x+\tan(\frac{9}{2}-\frac{17x}{2})+\frac{3}{2}$. The derivative of $-2x$ is $-2$. For the term $\tan(\frac{9}{2}-\frac{17x}{2})$, let $u=\frac{9}{2}-\frac{17x}{2}$. Then the derivative of $\tan u$ is $\sec^2 u$. The derivative of $u$ with respect to $x$ is $-\frac{17}{2}$. So the derivative of $\tan(\frac{9}{2}-\frac{17x}{2})$ is $-\frac{17}{2}\sec^2(\frac{9}{2}-\frac{17x}{2})$. So the derivative of $f(x)$ is $-2-\frac{17}{2}\sec^2(\frac{9}{2}-\frac{17x}{2})$. $\boxed{-2-\frac{17}{2}\sec^2(\frac{9}{2}-\frac{17x}{2})}$. |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| gpto1 | To find \( f'(x) \) for the function \( f(x) = -2x + \tan\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) + \dfrac{3}{2} \), we'll proceed step by step. **Step 1: Identify the inner function and its derivative.** Let \[ u(x) = \dfrac{9}{2} - \dfrac{17x}{2} \] Then \[ u'(x) = -\dfrac{17}{2} \] **Step 2: Differentiate each term of \( f(x) \) separately.** The derivative of \( -2x \) is \( -2 \). The derivative of \( \dfrac{3}{2} \) is \( 0 \) (since it's a constant). For the term involving the tangent function, use the chain rule: \[ \frac{d}{dx} \left[ \tan(u(x)) \right] = u'(x) \sec^2(u(x)) \] So, \[ \frac{d}{dx} \left[ \tan\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) \right] = -\dfrac{17}{2} \sec^2\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) \] **Step 3: Combine the derivatives.** Adding up the derivatives: \[ f'(x) = -2 - \dfrac{17}{2} \sec^2\left( \dfrac{9}{2} - \dfrac{17x}{2} \right) \] **Final Answer:** \[ \boxed{\, -2\ -\ \dfrac{17}{2}\ \sec^{2}\left(\dfrac{9}{2}-\dfrac{17x}{2}\right) \,} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| kimi | To differentiate the function \( f(x) = -2x + \tan \left( \frac{9}{2} - \frac{17x}{2} \right) + \frac{3}{2} \), we will differentiate each term separately and then combine the results. 1. **Differentiate \(-2x\):** The derivative of \(-2x\) with respect to \(x\) is: \[ \frac{d}{dx}(-2x) = -2 \] 2. **Differentiate \(\tan \left( \frac{9}{2} - \frac{17x}{2} \right)\):** We use the chain rule here. Let \( u = \frac{9}{2} - \frac{17x}{2} \). Then, \( \tan(u) \) and we need to find \( \frac{d}{dx} \tan(u) \). First, find the derivative of \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx} \left( \frac{9}{2} - \frac{17x}{2} \right) = 0 - \frac{17}{2} = -\frac{17}{2} \] Next, find the derivative of \( \tan(u) \) with respect to \( u \): \[ \frac{d}{du} \tan(u) = \sec^2(u) \] Now, apply the chain rule: \[ \frac{d}{dx} \tan \left( \frac{9}{2} - \frac{17x}{2} \right) = \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right) \cdot \left( -\frac{17}{2} \right) = -\frac{17}{2} \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right) \] 3. **Differentiate \(\frac{3}{2}\):** The derivative of a constant is zero: \[ \frac{d}{dx} \left( \frac{3}{2} \right) = 0 \] Now, combine the derivatives of all the terms: \[ f'(x) = -2 + \left( -\frac{17}{2} \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right) \right) + 0 \] Simplify the expression: \[ f'(x) = -2 - \frac{17}{2} \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right) \] Thus, the final answer is: \[ \boxed{-2 - \frac{17}{2} \sec^2 \left( \frac{9}{2} - \frac{17x}{2} \right)} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| minimax | To differentiate the function $f(x) = -2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2}$, we will apply the chain rule to the tangent function and the power rule to the linear term. The derivative of $-2x$ with respect to $x$ is $-2$. For the tangent function, we have to apply the chain rule. Let $u = \frac{9}{2} - \frac{17x}{2}$. Then the derivative of $\tan(u)$ with respect to $u$ is $\sec^2(u)$, and the derivative of $u$ with respect to $x$ is $-\frac{17}{2}$. So, the derivative of $\tan\left(\frac{9}{2} - \frac{17x}{2}\right)$ with respect to $x$ is $\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right)$. The derivative of the constant term $\frac{3}{2}$ with respect |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| wenxinyiyan | The derivative of the given function can be found by differentiating each term separately. For the first term, $-2x$, the derivative with respect to $x$ is simply $-2$. For the second term, $\tan\left(\frac{9}{2} - \frac{17x}{2}\right)$, we need to use the chain rule. The derivative of $\tan(u)$ with respect to $u$ is $\sec^2(u)$, and the derivative of $\frac{9}{2} - \frac{17x}{2}$ with respect to $x$ is $-\frac{17}{2}$. So, by the chain rule, the derivative of $\tan\left(\frac{9}{2} - \frac{17x}{2}\right)$ with respect to $x$ is $\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right)$. The third term, $\frac{3}{2}$, is a constant, so its derivative with respect to $x$ is 0. Putting these together, the derivative of the given function is: $\boxed{-2 - \frac{17}{2}\sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)}$ |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| zhipu | To differentiate the function \( f(x) = -2x + \tan\left(\frac{9}{2} - \frac{17x}{2}\right) + \frac{3}{2} \), we will differentiate each term separately. 1. **Differentiate \( -2x \):** \[ \frac{d}{dx}(-2x) = -2 \] 2. **Differentiate \( \tan\left(\frac{9}{2} - \frac{17x}{2}\right) \):** Let \( u = \frac{9}{2} - \frac{17x}{2} \). Then the function becomes \( \tan(u) \). The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \), and we need to multiply by the derivative of \( u \) with respect to \( x \): \[ \frac{d}{dx} \left( \tan(u) \right) = \sec^2(u) \cdot \frac{du}{dx} \] Now, compute \( \frac{du}{dx} \): \[ u = \frac{9}{2} - \frac{17x}{2} \implies \frac{du}{dx} = -\frac{17}{2} \] Therefore, \[ \frac{d}{dx} \left( \tan\left(\frac{9}{2} - \frac{17x}{2}\right) \right) = \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \cdot \left(-\frac{17}{2}\right) = -\frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) \] 3. **Differentiate \( \frac{3}{2} \):** \[ \frac{d}{dx} \left( \frac{3}{2} \right) = 0 \] Now, combine all the derivatives: \[ f'(x) = -2 - \frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right) + 0 \] Thus, the derivative of the function is: \[ \boxed{-2 - \frac{17}{2} \sec^2\left(\frac{9}{2} - \frac{17x}{2}\right)} \] |
-\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-db6a290f-0604-9884-bb46-a451b088c0ea', 'request_id': 'db6a290f-0604-9884-bb46-a451b088c0ea'} | -\frac{17}{2} \sec ^2\left(\frac{1}{2} (9-17 x)\right)-2 |