| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the function \(\sin \left(7 x^4+4\right) \cos (9-x)\), we will use the product rule and the chain rule. Let's denote the function as \(f(x) = \sin \left(7 x^4+4\right) \cos (9-x)\). The product rule states that if \(f(x) = u(x) v(x)\), then \(f'(x) = u'(x) v(x) + u(x) v'(x)\). Here, we let: \[ u(x) = \sin \left(7 x^4+4\right) \] \[ v(x) = \cos (9-x) \] First, we need to find the derivatives of \(u(x)\) and \(v(x)\). 1. **Derivative of \(u(x) = \sin \left(7 x^4+4\right)\):** Using the chain rule, we have: \[ u'(x) = \cos \left(7 x^4+4\right) \cdot \frac{d}{dx} \left(7 x^4+4\right) \] \[ \frac{d}{dx} \left(7 x^4+4\right) = 28 x^3 \] Therefore: \[ u'(x) = \cos \left(7 x^4+4\right) \cdot 28 x^3 \] 2. **Derivative of \(v(x) = \cos (9-x)\):** Using the chain rule, we have: \[ v'(x) = -\sin (9-x) \cdot \frac{d}{dx} (9-x) \] \[ \frac{d}{dx} (9-x) = -1 \] Therefore: \[ v'(x) = -\sin (9-x) \cdot (-1) = \sin (9-x) \] Now, applying the product rule: \[ f'(x) = u'(x) v(x) + u(x) v'(x) \] \[ f'(x) = \left( \cos \left(7 x^4+4\right) \cdot 28 x^3 \right) \cos (9-x) + \sin \left(7 x^4+4\right) \cdot \sin (9-x) \] Thus, the final differentiated function is: \[ f'(x) = 28 x^3 \cos \left(7 x^4+4\right) \cos (9-x) + \sin \left(7 x^4+4\right) \sin (9-x) \] Putting the final answer in a box: \[ \boxed{28 x^3 \cos \left(7 x^4+4\right) \cos (9-x) + \sin \left(7 x^4+4\right) \sin (9-x)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| deepseek-r1 | To differentiate the function \(\sin(7x^4 + 4) \cos(9 - x)\), we use the product rule. Let \(f(x) = \sin(7x^4 + 4)\) and \(g(x) = \cos(9 - x)\). The product rule states that the derivative \(h'(x)\) is given by \(f'(x)g(x) + f(x)g'(x)\). First, we find the derivative of \(f(x)\): \[ f'(x) = \cos(7x^4 + 4) \cdot \frac{d}{dx}(7x^4 + 4) = \cos(7x^4 + 4) \cdot 28x^3 = 28x^3 \cos(7x^4 + 4) \] Next, we find the derivative of \(g(x)\): \[ g'(x) = -\sin(9 - x) \cdot \frac{d}{dx}(9 - x) = -\sin(9 - x) \cdot (-1) = \sin(9 - x) \] Applying the product rule, we get: \[ h'(x) = f'(x)g(x) + f(x)g'(x) \] Substituting the derivatives we found: \[ h'(x) = 28x^3 \cos(7x^4 + 4) \cdot \cos(9 - x) + \sin(7x^4 + 4) \cdot \sin(9 - x) \] Thus, the derivative of the function is: \[ \boxed{28x^3 \cos(7x^4 + 4) \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| gpto1 | To find the derivative of the function \( f(x) = \sin(7x^4 + 4) \cos(9 - x) \), we'll use the product rule and trigonometric identities. **Steps:** 1. **Apply the Product Rule**: The product rule states that for two functions \( u(x) \) and \( v(x) \): \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] Here, \( u(x) = \sin(7x^4 + 4) \) and \( v(x) = \cos(9 - x) \). 2. **Find the Derivatives of \( u(x) \) and \( v(x) \)**: - **Derivative of \( u(x) \)**: \[ u'(x) = \cos(7x^4 + 4) \cdot (7x^4 + 4)' \] \[ (7x^4 + 4)' = 28x^3 \] Thus: \[ u'(x) = 28x^3 \cos(7x^4 + 4) \] - **Derivative of \( v(x) \)**: \[ v'(x) = -\sin(9 - x) \cdot (9 - x)' \] \[ (9 - x)' = -1 \] Thus: \[ v'(x) = -\sin(9 - x) \cdot (-1) = \sin(9 - x) \] 3. **Apply the Product Rule**: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Plug in \( u'(x) \), \( u(x) \), \( v'(x) \), and \( v(x) \): \[ f'(x) = \left[28x^3 \cos(7x^4 + 4)\right] \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x) \] 4. **Simplify Using Trigonometric Identities**: Use the identity: \[ \cos A \cos B + \sin A \sin B = \cos(A - B) \] Apply this identity: \[ f'(x) = 28x^3 \cos\left((7x^4 + 4) - (9 - x)\right) \] Simplify the argument: \[ (7x^4 + 4) - (9 - x) = 7x^4 + x - 5 \] So: \[ f'(x) = 28x^3 \cos(7x^4 + x - 5) \] **Final Answer:** \[ \boxed{28x^3\,\cos\left(7x^4+x-5\right)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| gpt-4o-2024-08-06 | To differentiate the function \( f(x) = \sin(7x^4 + 4) \cos(9 - x) \), we will use the product rule and the chain rule. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product is given by: \[ (uv)' = u'v + uv' \] Let \( u(x) = \sin(7x^4 + 4) \) and \( v(x) = \cos(9 - x) \). We need to find \( u'(x) \) and \( v'(x) \). **Step 1: Differentiate \( u(x) = \sin(7x^4 + 4) \).** Using the chain rule, the derivative of \( \sin(g(x)) \) is \( \cos(g(x)) \cdot g'(x) \). Here, \( g(x) = 7x^4 + 4 \), so: \[ g'(x) = \frac{d}{dx}(7x^4 + 4) = 28x^3 \] Thus, the derivative of \( u(x) \) is: \[ u'(x) = \cos(7x^4 + 4) \cdot 28x^3 = 28x^3 \cos(7x^4 + 4) \] **Step 2: Differentiate \( v(x) = \cos(9 - x) \).** Again using the chain rule, the derivative of \( \cos(h(x)) \) is \(-\sin(h(x)) \cdot h'(x)\). Here, \( h(x) = 9 - x \), so: \[ h'(x) = \frac{d}{dx}(9 - x) = -1 \] Thus, the derivative of \( v(x) \) is: \[ v'(x) = -\sin(9 - x) \cdot (-1) = \sin(9 - x) \] **Step 3: Apply the product rule.** Now, using the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substitute the expressions for \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \): \[ f'(x) = \left(28x^3 \cos(7x^4 + 4)\right) \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x) \] Simplifying, we have: \[ f'(x) = 28x^3 \cos(7x^4 + 4) \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x) \] Thus, the derivative of the function is: \[ \boxed{28x^3 \cos(7x^4 + 4) \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| DeepSeek-Coder-V2-Lite-Instruct | To differentiate the function \(\sin \left(7 x^4+4\right) \cos (9-x)\), we will use the product rule. The product rule states that if you have a product of two functions \(u(x)\) and \(v(x)\), the derivative of their product is given by: \[ (uv)' = u'v + uv' \] Here, let: \[ u(x) = \sin \left(7 x^4 + 4\right) \] \[ v(x) = \cos (9 - x) \] First, we need to find the derivatives of \(u(x)\) and \(v(x)\). 1. **Differentiate \(u(x)\):** \[ u(x) = \sin \left(7 x^4 + 4\right) \] Using the chain rule, the derivative of \(u(x)\) is: \[ u'(x) = \cos \left(7 x^4 + 4\right) \cdot \frac{d}{dx} \left(7 x^4 + 4\right) \] \[ \frac{d}{dx} \left(7 x^4 + 4\right) = 28 x^3 \] So, \[ u'(x) = 28 x^3 \cos \left(7 x^4 + 4\right) \] 2. **Differentiate \(v(x)\):** \[ v(x) = \cos (9 - x) \] Using the chain rule, the derivative of \(v(x)\) is: \[ v'(x) = -\sin (9 - x) \cdot \frac{d}{dx} (9 - x) \] \[ \frac{d}{dx} (9 - x) = -1 \] So, \[ v'(x) = -\sin (9 - x) \] Now, applying the product rule: \[ \frac{d}{dx} \left[\sin \left(7 x^4 + 4\right) \cos (9 - x)\right] = u'(x)v(x) + u(x)v'(x) \] Substituting the derivatives we found: \[ \frac{d}{dx} \left[\sin \left(7 x^4 + 4\right) \cos (9 - x)\right] = \left(28 x^3 \cos \left(7 x^4 + 4\right)\right) \cos (9 - x) + \sin \left(7 x^4 + 4\right) \left(-\sin (9 - x)\right) \] Simplifying, we get: \[ \frac{d}{dx} \left[\sin \left(7 x^4 + 4\right) \cos (9 - x)\right] = 28 x^3 \cos \left(7 x^4 + 4\right) \cos (9 - x) - \sin \left(7 x^4 + 4\right) \sin (9 - x) \] Thus, the final answer is: \[ \boxed{28 x^3 \cos \left(7 x^4 + 4\right) \cos (9 - x) - \sin \left(7 x^4 + 4\right) \sin (9 - x)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $\sin \left(7 x^4+4\right) \cos (9-x)$. This is a product of two functions, so we will need to use the product rule for differentiation. ## Step 2: Apply the product rule for differentiation The product rule states that if we have a function of the form $f(x)g(x)$, then its derivative is given by $f'(x)g(x) + f(x)g'(x)$. In this case, $f(x) = \sin \left(7 x^4+4\right)$ and $g(x) = \cos (9-x)$. ## Step 3: Differentiate the first function, $f(x) = \sin \left(7 x^4+4\right)$ To differentiate $\sin \left(7 x^4+4\right)$, we use the chain rule. The derivative of $\sin(u)$ is $\cos(u)$, and the derivative of $7x^4+4$ is $28x^3$. Therefore, the derivative of $\sin \left(7 x^4+4\right)$ is $\cos \left(7 x^4+4\right) \cdot 28x^3$. ## Step 4: Differentiate the second function, $g(x) = \cos (9-x)$ To differentiate $\cos (9-x)$, we use the chain rule. The derivative of $\cos(u)$ is $-\sin(u)$, and the derivative of $9-x$ is $-1$. Therefore, the derivative of $\cos (9-x)$ is $-\sin (9-x) \cdot -1 = \sin (9-x)$. ## Step 5: Combine the derivatives using the product rule Applying the product rule, the derivative of the given function is $\cos \left(7 x^4+4\right) \cdot 28x^3 \cdot \cos (9-x) + \sin \left(7 x^4+4\right) \cdot \sin (9-x)$. ## Step 6: Simplify the expression if necessary No further simplification is needed. The final answer is: $\boxed{28x^3 \cos \left(7 x^4+4\right) \cos (9-x) + \sin \left(7 x^4+4\right) \sin (9-x)}$ |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $\sin \left(7 x^4+4\right) \cos (9-x)$. This is a product of two functions, so we will need to use the product rule of differentiation. ## Step 2: Apply the product rule of differentiation The product rule states that if we have a function of the form $u(x)v(x)$, then the derivative is given by $u'(x)v(x) + u(x)v'(x)$. In this case, $u(x) = \sin \left(7 x^4+4\right)$ and $v(x) = \cos (9-x)$. ## Step 3: Differentiate the first function, u(x) To differentiate $u(x) = \sin \left(7 x^4+4\right)$, we will use the chain rule. The derivative of $\sin$ is $\cos$, and the derivative of $7x^4+4$ is $28x^3$. So, $u'(x) = \cos \left(7 x^4+4\right) \cdot 28x^3$. ## Step 4: Differentiate the second function, v(x) To differentiate $v(x) = \cos (9-x)$, we will also use the chain rule. The derivative of $\cos$ is $-\sin$, and the derivative of $9-x$ is $-1$. So, $v'(x) = -\sin (9-x) \cdot -1 = \sin (9-x)$. ## Step 5: Combine the derivatives using the product rule Now we can combine the derivatives using the product rule: $u'(x)v(x) + u(x)v'(x) = \cos \left(7 x^4+4\right) \cdot 28x^3 \cdot \cos (9-x) + \sin \left(7 x^4+4\right) \cdot \sin (9-x)$. ## Step 6: Simplify the expression The expression can be simplified by combining like terms, but in this case, it is already in a relatively simple form. The final answer is: $\boxed{28x^3 \cos \left(7 x^4+4\right) \cos (9-x) + \sin \left(7 x^4+4\right) \sin (9-x)}$ |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| Qwen2-72B-Instruct | To differentiate the given function, we will use the product rule and the chain rule. The function is: \[f(x) = \sin \left(7 x^4 + 4\right) \cos (9 - x)\] Let's denote: \[u(x) = \sin \left(7 x^4 + 4\right)\] \[v(x) = \cos (9 - x)\] The derivative of $f(x)$ can be found using the product rule: \[f'(x) = u'(x)v(x) + u(x)v'(x)\] First, we find the derivatives of $u(x)$ and $v(x)$. For $u(x)$: \[u'(x) = \frac{d}{dx} \sin \left(7 x^4 + 4\right)\] Using the chain rule: \[u'(x) = \cos \left(7 x^4 + 4\right) \cdot \frac{d}{dx} \left(7 x^4 + 4\right)\] \[u'(x) = \cos \left(7 x^4 + 4\right) \cdot 28x^3\] For $v(x)$: \[v'(x) = \frac{d}{dx} \cos (9 - x)\] Using the chain rule: \[v'(x) = -\sin (9 - x) \cdot \frac{d}{dx} (9 - x)\] \[v'(x) = -\sin (9 - x) \cdot (-1)\] \[v'(x) = \sin (9 - x)\] Now, we apply the product rule: \[f'(x) = u'(x)v(x) + u(x)v'(x)\] \[f'(x) = \left(\cos \left(7 x^4 + 4\right) \cdot 28x^3\right) \cos (9 - x) + \sin \left(7 x^4 + 4\right) \sin (9 - x)\] Therefore, the derivative of the function is: \[f'(x) = 28x^3\cos \left(7 x^4 + 4\right) \cos (9 - x) + \sin \left(7 x^4 + 4\right) \sin (9 - x)\] In $\boxed{}$ format: \[ \boxed{f'(x) = 28x^3\cos \left(7 x^4 + 4\right) \cos (9 - x) + \sin \left(7 x^4 + 4\right) \sin (9 - x)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| baichuan4 | To differentiate the function $\sin \left(7 x^4+4\right) \cos (9-x)$, we will use the product rule and the chain rule. The product rule states that if $u(x)$ and $v(x)$ are two functions, then the derivative of their product is given by $(uv)' = u'v + uv'$. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}$. Let's start by applying the product rule: $$ \frac{d}{dx}[\sin(7x^4+4)\cos(9-x)] = \frac{d}{dx}[\sin(7x^4+4)]\cos(9-x) + \sin(7x^4+4)\frac{d}{dx}[\cos(9-x)] $$ Now we need to apply the chain rule to differentiate $\sin(7x^4+4)$ and $\cos(9-x)$. For $\sin(7x^4+4)$, let $u = 7x^4+4$. Then $\frac{du}{dx} = 28x^3$. By the chain rule, we have: $$ \frac{d}{dx}[\sin(7x^4+4)] = \cos(7x^4+4) \cdot \frac{du}{dx} = \cos(7x^4+4) \cdot 28x^3 $$ For $\cos(9-x)$, let $v = 9-x$. Then $\frac{dv}{dx} = -1$. By the chain rule, we have: $$ \frac{d}{dx}[\cos(9-x)] = -\sin(9-x) \cdot \frac{dv}{dx} = -\sin(9-x) \cdot (-1) = \sin(9-x) $$ Now we can substitute these derivatives back into the product rule: $$ \frac{d}{dx}[\sin(7x^4+4)\cos(9-x)] = (\cos(7x^4+4) \cdot 28x^3)\cos(9-x) + \sin(7x^4+4)(\sin(9-x)) $$ Simplifying, we get: $$ \frac{d}{dx}[\sin(7x^4+4)\cos(9-x)] = 28x^3\cos(7x^4+4)\cos(9-x) + \sin(7x^4+4)\sin(9-x) $$ This is the final derivative of the function. We can put it in a box as requested: $$ \boxed{28x^3\cos(7x^4+4)\cos(9-x) + \sin(7x^4+4)\sin(9-x)} $$ |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| doubao | Let $y=\sin(7x^4+4)\cos(9-x)$. Using the product rule, the derivative is: \[ \begin{align*} y'&=(\sin(7x^4+4))'\cos(9-x)+\sin(7x^4+4)(\cos(9-x))'\\ &=(28x^3\cos(7x^4+4))\cos(9-x)-\sin(7x^4+4)\sin(9-x)\\ &=\boxed{28x^3\cos(7x^4+4)\cos(9-x)-\sin(7x^4+4)\sin(9-x)} \end{align*} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| gpto1 | To find the derivative of the function \( f(x) = \sin(7x^4 + 4) \cos(9 - x) \), we'll use the product rule and trigonometric identities. **Steps:** 1. **Apply the Product Rule**: The product rule states that for two functions \( u(x) \) and \( v(x) \): \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] Here, \( u(x) = \sin(7x^4 + 4) \) and \( v(x) = \cos(9 - x) \). 2. **Find the Derivatives of \( u(x) \) and \( v(x) \)**: - **Derivative of \( u(x) \)**: \[ u'(x) = \cos(7x^4 + 4) \cdot (7x^4 + 4)' \] \[ (7x^4 + 4)' = 28x^3 \] Thus: \[ u'(x) = 28x^3 \cos(7x^4 + 4) \] - **Derivative of \( v(x) \)**: \[ v'(x) = -\sin(9 - x) \cdot (9 - x)' \] \[ (9 - x)' = -1 \] Thus: \[ v'(x) = -\sin(9 - x) \cdot (-1) = \sin(9 - x) \] 3. **Apply the Product Rule**: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Plug in \( u'(x) \), \( u(x) \), \( v'(x) \), and \( v(x) \): \[ f'(x) = \left[28x^3 \cos(7x^4 + 4)\right] \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x) \] 4. **Simplify Using Trigonometric Identities**: Use the identity: \[ \cos A \cos B + \sin A \sin B = \cos(A - B) \] Apply this identity: \[ f'(x) = 28x^3 \cos\left((7x^4 + 4) - (9 - x)\right) \] Simplify the argument: \[ (7x^4 + 4) - (9 - x) = 7x^4 + x - 5 \] So: \[ f'(x) = 28x^3 \cos(7x^4 + x - 5) \] **Final Answer:** \[ \boxed{28x^3\,\cos\left(7x^4+x-5\right)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| kimi | To differentiate the function \( f(x) = \sin(7x^4 + 4) \cos(9 - x) \), we will use the product rule and the chain rule. The product rule states that if \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \). Let \( u(x) = \sin(7x^4 + 4) \) and \( v(x) = \cos(9 - x) \). First, we find the derivatives of \( u(x) \) and \( v(x) \). 1. **Differentiate \( u(x) = \sin(7x^4 + 4) \):** - Use the chain rule: \( \frac{d}{dx}[\sin(g(x))] = \cos(g(x)) \cdot g'(x) \). - Here, \( g(x) = 7x^4 + 4 \). - Compute \( g'(x) \): \[ g'(x) = \frac{d}{dx}[7x^4 + 4] = 28x^3 \] - Therefore, \[ u'(x) = \cos(7x^4 + 4) \cdot 28x^3 = 28x^3 \cos(7x^4 + 4) \] 2. **Differentiate \( v(x) = \cos(9 - x) \):** - Use the chain rule: \( \frac{d}{dx}[\cos(h(x))] = -\sin(h(x)) \cdot h'(x) \). - Here, \( h(x) = 9 - x \). - Compute \( h'(x) \): \[ h'(x) = \frac{d}{dx}[9 - x] = -1 \] - Therefore, \[ v'(x) = -\sin(9 - x) \cdot (-1) = \sin(9 - x) \] Now, apply the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ f'(x) = (28x^3 \cos(7x^4 + 4)) \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x) \] Thus, the final answer is: \[ \boxed{28x^3 \cos(7x^4 + 4) \cos(9 - x) + \sin(7x^4 + 4) \sin(9 - x)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| minimax | To differentiate the function $f(x) = \sin(7x^4 + 4) \cos(9 - x)$, we will use the product rule and the chain rule. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. The chain rule is used to differentiate composite functions. Let's denote $u(x) = \sin(7x^4 + 4)$ and $v(x) = \cos(9 - x)$. We need to find $u'(x)$ and $v'(x)$. First, differentiate $u(x)$ using the chain rule: $$ u'(x) = \frac{d}{dx}[\sin(7x^4 + 4)] = \cos(7x^4 + 4) \cdot \frac{d}{dx}[7x^4 + 4] = \cos(7x^4 + 4) \cdot 28x^3. $$ Next, differentiate $v(x)$: $$ v'(x) = \frac{d}{dx}[\cos(9 - x)] = -\sin(9 |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| wenxinyiyan | To differentiate the given function, we apply the product rule, which states that if $u$ and $v$ are functions of $x$, then the derivative of their product $uv$ is given by $u'v + uv'$. Let $u = \sin(7x^4 + 4)$ and $v = \cos(9 - x)$. First, we find the derivative of $u$ with respect to $x$: $$u' = \frac{d}{dx}\sin(7x^4 + 4) = \cos(7x^4 + 4) \cdot \frac{d}{dx}(7x^4 + 4) = 28x^3 \cos(7x^4 + 4).$$ Next, we find the derivative of $v$ with respect to $x$: $$v' = \frac{d}{dx}\cos(9 - x) = -\sin(9 - x) \cdot \frac{d}{dx}(9 - x) = \sin(9 - x).$$ Now, applying the product rule: $$\frac{d}{dx}(\sin(7x^4 + 4)\cos(9 - x)) = u'v + uv' = 28x^3 \cos(7x^4 + 4)\cos(9 - x) + \sin(7x^4 + 4)\sin(9 - x).$$ Thus, the final answer is: $$\boxed{28x^3 \cos(7x^4 + 4)\cos(9 - x) + \sin(7x^4 + 4)\sin(9 - x)}.$$ |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| zhipu | To differentiate the function \( f(x) = \sin \left(7 x^4 + 4\right) \cos (9 - x) \), we will use the product rule and the chain rule. The product rule states that if you have a function \( u(x) v(x) \), its derivative is given by: \[ (uv)' = u'v + uv'. \] Let: \[ u(x) = \sin \left(7 x^4 + 4\right) \] \[ v(x) = \cos (9 - x) \] First, we find the derivatives of \( u(x) \) and \( v(x) \): 1. **Derivative of \( u(x) \):** \[ u(x) = \sin \left(7 x^4 + 4\right) \] Using the chain rule, let \( g(x) = 7 x^4 + 4 \) and \( h(g) = \sin(g) \). Then: \[ u'(x) = h'(g) \cdot g'(x) \] \[ h'(g) = \cos(g) \] \[ g'(x) = 28 x^3 \] So: \[ u'(x) = \cos \left(7 x^4 + 4\right) \cdot 28 x^3 \] 2. **Derivative of \( v(x) \):** \[ v(x) = \cos (9 - x) \] Using the chain rule, let \( k(x) = 9 - x \) and \( m(k) = \cos(k) \). Then: \[ v'(x) = m'(k) \cdot k'(x) \] \[ m'(k) = -\sin(k) \] \[ k'(x) = -1 \] So: \[ v'(x) = -\sin (9 - x) \cdot (-1) = \sin (9 - x) \] Now, applying the product rule: \[ f'(x) = u'(x) v(x) + u(x) v'(x) \] \[ f'(x) = \left( \cos \left(7 x^4 + 4\right) \cdot 28 x^3 \right) \cos (9 - x) + \sin \left(7 x^4 + 4\right) \left( \sin (9 - x) \right) \] Thus, the derivative of the function is: \[ \boxed{28 x^3 \cos \left(7 x^4 + 4\right) \cos (9 - x) + \sin \left(7 x^4 + 4\right) \sin (9 - x)} \] |
\sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-4fdf33a0-c915-9b16-8ffb-dc13795255c3', 'request_id': '4fdf33a0-c915-9b16-8ffb-dc13795255c3'} | \sin (9-x) \sin \left(7 x^4+4\right)+28 x^3 \cos (9-x) \cos \left(7 x^4+4\right) |