| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the function \( e^{3 x^5-6} \tan \left(x^2+9\right) \), we will use the product rule and the chain rule. Let's break it down step by step. First, let's define the two parts of the function: \[ u(x) = e^{3 x^5-6} \] \[ v(x) = \tan \left(x^2+9\right) \] The product rule states that: \[ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \] We need to find the derivatives \( u'(x) \) and \( v'(x) \). ### Step 1: Differentiate \( u(x) = e^{3 x^5-6} \) Using the chain rule: \[ u'(x) = e^{3 x^5-6} \cdot \frac{d}{dx} (3 x^5 - 6) \] \[ \frac{d}{dx} (3 x^5 - 6) = 15 x^4 \] So, \[ u'(x) = 15 x^4 e^{3 x^5-6} \] ### Step 2: Differentiate \( v(x) = \tan \left(x^2+9\right) \) Using the chain rule: \[ v'(x) = \sec^2 \left(x^2+9\right) \cdot \frac{d}{dx} (x^2 + 9) \] \[ \frac{d}{dx} (x^2 + 9) = 2 x \] So, \[ v'(x) = 2 x \sec^2 \left(x^2+9\right) \] ### Step 3: Apply the product rule Now, we combine the derivatives using the product rule: \[ \frac{d}{dx} [e^{3 x^5-6} \tan \left(x^2+9\right)] = u'(x) v(x) + u(x) v'(x) \] \[ = (15 x^4 e^{3 x^5-6}) \tan \left(x^2+9\right) + e^{3 x^5-6} (2 x \sec^2 \left(x^2+9\right)) \] ### Step 4: Simplify the expression Combine the terms: \[ \frac{d}{dx} [e^{3 x^5-6} \tan \left(x^2+9\right)] = 15 x^4 e^{3 x^5-6} \tan \left(x^2+9\right) + 2 x e^{3 x^5-6} \sec^2 \left(x^2+9\right) \] Thus, the final differentiated function is: \[ \boxed{15 x^4 e^{3 x^5-6} \tan \left(x^2+9\right) + 2 x e^{3 x^5-6} \sec^2 \left(x^2+9\right)} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| deepseek-r1 | To differentiate the function \( e^{3x^5 - 6} \tan(x^2 + 9) \), we use the product rule. Let \( u(x) = e^{3x^5 - 6} \) and \( v(x) = \tan(x^2 + 9) \). The product rule states that the derivative is \( u'(x)v(x) + u(x)v'(x) \). First, we find the derivatives of \( u(x) \) and \( v(x) \): 1. The derivative of \( u(x) = e^{3x^5 - 6} \) using the chain rule: \[ u'(x) = e^{3x^5 - 6} \cdot \frac{d}{dx}(3x^5 - 6) = e^{3x^5 - 6} \cdot 15x^4 \] 2. The derivative of \( v(x) = \tan(x^2 + 9) \) using the chain rule: \[ v'(x) = \sec^2(x^2 + 9) \cdot \frac{d}{dx}(x^2 + 9) = \sec^2(x^2 + 9) \cdot 2x \] Applying the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substituting the derivatives we found: \[ f'(x) = [15x^4 e^{3x^5 - 6}] \tan(x^2 + 9) + e^{3x^5 - 6} [2x \sec^2(x^2 + 9)] \] Factoring out the common terms \( e^{3x^5 - 6} \) and \( x \): \[ f'(x) = e^{3x^5 - 6} \cdot x \left(15x^3 \tan(x^2 + 9) + 2 \sec^2(x^2 + 9)\right) \] Thus, the derivative of the function is: \[ \boxed{e^{3x^5 -6} x \left(15x^3 \tan(x^2 + 9) + 2 \sec^2(x^2 + 9)\right)} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| gpto1 | To differentiate the function \( f(x) = e^{3 x^5 - 6} \tan(x^2 + 9) \), we will use the product rule: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] where \( u(x) = e^{3 x^5 -6} \) and \( v(x) = \tan(x^2 + 9) \). First, find \( u'(x) \): \[ u'(x) = e^{3 x^5 -6} \cdot \frac{d}{dx}(3 x^5 -6) = e^{3 x^5 -6} \cdot 15 x^4 \] Next, find \( v'(x) \): \[ v'(x) = \sec^2(x^2 + 9) \cdot \frac{d}{dx}(x^2 + 9) = \sec^2(x^2 + 9) \cdot 2x \] Now apply the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) = \left( e^{3 x^5 -6} \cdot 15 x^4 \right) \tan(x^2 + 9) + e^{3 x^5 -6} \cdot \left( \sec^2(x^2 + 9) \cdot 2x \right) \] Simplify and factor out the common term \( e^{3 x^5 -6} \): \[ f'(x) = e^{3 x^5 -6} \left( 15 x^4 \tan(x^2 + 9) + 2x \sec^2(x^2 + 9) \right) \] So, the derivative of the given function is: \[ \boxed{e^{3 x^5 -6}\left(15x^{4}\tan\left(x^{2}+9\right)+2x\sec^{2}\left(x^{2}+9\right)\right)} \] \(\boxed{e^{3 x^{5}-6}\left(15x^{4}\tan(x^{2}+9)+2x\,\sec^{2}(x^{2}+9)\right)}\) |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| gpt-4o-2024-08-06 | To differentiate the function \( f(x) = e^{3x^5 - 6} \tan(x^2 + 9) \), 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 \( u(x) = e^{3x^5 - 6} \) and \( v(x) = \tan(x^2 + 9) \). First, we differentiate \( u(x) \): \[ u(x) = e^{3x^5 - 6} \] Using the chain rule, the derivative is: \[ u'(x) = e^{3x^5 - 6} \cdot \frac{d}{dx}(3x^5 - 6) \] \[ = e^{3x^5 - 6} \cdot (15x^4) \] \[ = 15x^4 e^{3x^5 - 6} \] Next, we differentiate \( v(x) \): \[ v(x) = \tan(x^2 + 9) \] Again, using the chain rule, the derivative is: \[ v'(x) = \sec^2(x^2 + 9) \cdot \frac{d}{dx}(x^2 + 9) \] \[ = \sec^2(x^2 + 9) \cdot (2x) \] \[ = 2x \sec^2(x^2 + 9) \] Now, apply the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] \[ = \left(15x^4 e^{3x^5 - 6}\right) \tan(x^2 + 9) + \left(e^{3x^5 - 6}\right) \left(2x \sec^2(x^2 + 9)\right) \] Simplify the expression: \[ = 15x^4 e^{3x^5 - 6} \tan(x^2 + 9) + 2x e^{3x^5 - 6} \sec^2(x^2 + 9) \] Factor out the common term \( e^{3x^5 - 6} \): \[ = e^{3x^5 - 6} \left( 15x^4 \tan(x^2 + 9) + 2x \sec^2(x^2 + 9) \right) \] Thus, the derivative of the function is: \[ \boxed{e^{3x^5 - 6} \left( 15x^4 \tan(x^2 + 9) + 2x \sec^2(x^2 + 9) \right)} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| DeepSeek-Coder-V2-Lite-Instruct | The differentiation of the function \( e^{3 x^5-6} \tan \left(x^2+9\right) \) can be found using the product rule, which 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. Let \( u = e^{3 x^5-6} \) and \( v = \tan \left(x^2+9\right) \). First, we find the derivative of \( u \): \[ u = e^{3 x^5-6} \] Using the chain rule: \[ \frac{d}{dx} e^{3 x^5-6} = e^{3 x^5-6} \cdot \frac{d}{dx} (3 x^5 - 6) = e^{3 x^5-6} \cdot (15 x^4) = 15 x^4 e^{3 x^5-6} \] Next, we find the derivative of \( v \): \[ v = \tan \left(x^2+9\right) \] Using the chain rule: \[ \frac{d}{dx} \tan \left(x^2+9\right) = \sec^2 \left(x^2+9\right) \cdot \frac{d}{dx} (x^2 + 9) = \sec^2 \left(x^2+9\right) \cdot (2 x) = 2 x \sec^2 \left(x^2+9\right) \] Now, applying the product rule: \[ \frac{d}{dx} \left( e^{3 x^5-6} \tan \left(x^2+9\right) \right) = u' v + u v' \] Substituting the derivatives we found: \[ \frac{d}{dx} \left( e^{3 x^5-6} \tan \left(x^2+9\right) \right) = (15 x^4 e^{3 x^5-6}) \tan \left(x^2+9\right) + e^{3 x^5-6} (2 x \sec^2 \left(x^2+9\right)) \] Combining the terms: \[ \frac{d}{dx} \left( e^{3 x^5-6} \tan \left(x^2+9\right) \right) = 15 x^4 e^{3 x^5-6} \tan \left(x^2+9\right) + 2 x e^{3 x^5-6} \sec^2 \left(x^2+9\right) \] Thus, the final answer is: \[ \boxed{15 x^4 e^{3 x^5-6} \tan \left(x^2+9\right) + 2 x e^{3 x^5-6} \sec^2 \left(x^2+9\right)} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $e^{3 x^5-6} \tan \left(x^2+9\right)$. This is a product of two functions, $e^{3 x^5-6}$ and $\tan \left(x^2+9\right)$, 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, $f(x) = e^{3 x^5-6}$ and $g(x) = \tan \left(x^2+9\right)$. ## Step 3: Differentiate the first function, $f(x) = e^{3 x^5-6}$ To differentiate $e^{3 x^5-6}$, we use the chain rule. The derivative of $e^u$ is $e^u \cdot u'$, where $u = 3 x^5-6$. So, $u' = 15x^4$, and the derivative of $f(x)$ is $f'(x) = e^{3 x^5-6} \cdot 15x^4$. ## Step 4: Differentiate the second function, $g(x) = \tan \left(x^2+9\right)$ To differentiate $\tan \left(x^2+9\right)$, we also use the chain rule. The derivative of $\tan(u)$ is $\sec^2(u) \cdot u'$, where $u = x^2+9$. So, $u' = 2x$, and the derivative of $g(x)$ is $g'(x) = \sec^2 \left(x^2+9\right) \cdot 2x$. ## Step 5: Combine the derivatives using the product rule Now, we combine the derivatives of $f(x)$ and $g(x)$ using the product rule: $f'(x)g(x) + f(x)g'(x) = (e^{3 x^5-6} \cdot 15x^4) \tan \left(x^2+9\right) + e^{3 x^5-6} (\sec^2 \left(x^2+9\right) \cdot 2x)$. ## Step 6: Simplify the expression Simplifying the expression gives us the derivative of the original function: $e^{3 x^5-6} (15x^4 \tan \left(x^2+9\right) + 2x \sec^2 \left(x^2+9\right))$. The final answer is: $\boxed{e^{3 x^5-6} (15x^4 \tan \left(x^2+9\right) + 2x \sec^2 \left(x^2+9\right))}$ |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $e^{3 x^5-6} \tan \left(x^2+9\right)$. This is a product of two functions, $e^{3 x^5-6}$ and $\tan \left(x^2+9\right)$. ## Step 2: Apply the product rule of differentiation To differentiate the given function, we will use the product rule of differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. ## Step 3: Differentiate the first function, $e^{3 x^5-6}$ Using the exponential function differentiation rule, which states that if $f(x) = e^{u(x)}$, then $f'(x) = e^{u(x)}u'(x)$, we can differentiate $e^{3 x^5-6}$. Here, $u(x) = 3x^5 - 6$, so $u'(x) = 15x^4$. Therefore, the derivative of $e^{3 x^5-6}$ is $e^{3 x^5-6} \cdot 15x^4$. ## Step 4: Differentiate the second function, $\tan \left(x^2+9\right)$ Using the chain rule of differentiation, which states that if $f(x) = g(h(x))$, then $f'(x) = g'(h(x))h'(x)$, we can differentiate $\tan \left(x^2+9\right)$. Here, $g(u) = \tan u$ and $h(x) = x^2 + 9$, so $g'(u) = \sec^2 u$ and $h'(x) = 2x$. Therefore, the derivative of $\tan \left(x^2+9\right)$ is $\sec^2 \left(x^2+9\right) \cdot 2x$. ## Step 5: Apply the product rule Now, we can apply the product rule by multiplying the derivative of the first function by the second function and adding the product of the first function and the derivative of the second function: $e^{3 x^5-6} \cdot 15x^4 \cdot \tan \left(x^2+9\right) + e^{3 x^5-6} \cdot \sec^2 \left(x^2+9\right) \cdot 2x$. ## Step 6: Simplify the expression Combining the terms, we get $e^{3 x^5-6} \left(15x^4 \tan \left(x^2+9\right) + 2x \sec^2 \left(x^2+9\right)\right)$. The final answer is: $\boxed{e^{3 x^5-6} \left(15x^4 \tan \left(x^2+9\right) + 2x \sec^2 \left(x^2+9\right)\right)}$ |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| Qwen2-72B-Instruct | To differentiate the given function, we will use the product rule and the chain rule. The function is: \[f(x) = e^{3x^5 - 6} \tan(x^2 + 9)\] Let's denote \(u(x) = e^{3x^5 - 6}\) and \(v(x) = \tan(x^2 + 9)\). Then, \[f'(x) = u'(x)v(x) + u(x)v'(x)\] First, let's find \(u'(x)\): \[u(x) = e^{3x^5 - 6}\] Using the chain rule: \[u'(x) = e^{3x^5 - 6} \cdot (15x^4)\] Next, let's find \(v'(x)\): \[v(x) = \tan(x^2 + 9)\] Again using the chain rule: \[v'(x) = \sec^2(x^2 + 9) \cdot (2x)\] Now, we can substitute these derivatives back into our expression for \(f'(x)\): \[f'(x) = [e^{3x^5 - 6} \cdot (15x^4)] \tan(x^2 + 9) + e^{3x^5 - 6} [\sec^2(x^2 + 9) \cdot (2x)]\] Simplifying this expression gives us the derivative of the original function: \[f'(x) = 15x^4e^{3x^5 - 6}\tan(x^2 + 9) + 2xe^{3x^5 - 6}\sec^2(x^2 + 9)\] Therefore, the final answer is: \[ \boxed{f'(x) = 15x^4e^{3x^5 - 6}\tan(x^2 + 9) + 2xe^{3x^5 - 6}\sec^2(x^2 + 9)} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| baichuan4 | To differentiate the function $f(x) = e^{3x^5-6} \tan(x^2+9)$, we will use the product rule and the chain rule. The product rule states that if $u(x)$ and $v(x)$ are functions, then the derivative of their product $u(x)v(x)$ is given by: $$ (uv)' = u'v + uv' $$ The chain rule states that if $g(x)$ is a function and $h(u)$ is a function of $u$, then the derivative of the composite function $h(g(x))$ is given by: $$ \frac{d}{dx}h(g(x)) = h'(g(x)) \cdot g'(x) $$ First, let's define our functions: $$ u(x) = e^{3x^5-6} $$ $$ v(x) = \tan(x^2+9) $$ Now, we will find the derivatives of $u(x)$ and $v(x)$ using the chain rule. For $u(x)$, we have: $$ u'(x) = \frac{d}{dx}e^{3x^5-6} $$ Using the chain rule, we get: $$ u'(x) = e^{3x^5-6} \cdot \frac{d}{dx}(3x^5-6) $$ $$ u'(x) = e^{3x^5-6} \cdot 15x^4 $$ For $v(x)$, we have: $$ v'(x) = \frac{d}{dx}\tan(x^2+9) $$ Using the chain rule, we get: $$ v'(x) = \sec^2(x^2+9) \cdot \frac{d}{dx}(x^2+9) $$ $$ v'(x) = \sec^2(x^2+9) \cdot 2x $$ Now, we apply the product rule to find the derivative of $f(x)$: $$ f'(x) = u'(x)v(x) + u(x)v'(x) $$ $$ f'(x) = (e^{3x^5-6} \cdot 15x^4) \tan(x^2+9) + e^{3x^5-6} (\sec^2(x^2+9) \cdot 2x) $$ Simplifying, we get: $$ f'(x) = 15x^4 e^{3x^5-6} \tan(x^2+9) + 2x e^{3x^5-6} \sec^2(x^2+9) $$ We can combine the terms with the common factor $e^{3x^5-6}$: $$ f'(x) = e^{3x^5-6} (15x^4 \tan(x^2+9) + 2x \sec^2(x^2+9)) $$ This is the final derivative of the function $f(x)$. We can put it in a box as requested: $$ \boxed{f'(x) = e^{3x^5-6} (15x^4 \tan(x^2+9) + 2x \sec^2(x^2+9))} $$ |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| doubao | Let $y=e^{3x^5-6}\tan(x^2+9)$. Using the product rule and the chain rule: \[ \begin{align*} y'&=(e^{3x^5-6})'\tan(x^2+9)+e^{3x^5-6}(\tan(x^2+9))'\\ &=(3\times5x^4e^{3x^5-6})\tan(x^2+9)+e^{3x^5-6}(\sec^2(x^2+9))\times 2x\\ &=\boxed{15x^4e^{3x^5-6}\tan(x^2+9)+2xe^{3x^5-6}\sec^2(x^2+9)} \end{align*} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| gpto1 | To differentiate the function \( f(x) = e^{3 x^5 - 6} \tan(x^2 + 9) \), we will use the product rule: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] where \( u(x) = e^{3 x^5 -6} \) and \( v(x) = \tan(x^2 + 9) \). First, find \( u'(x) \): \[ u'(x) = e^{3 x^5 -6} \cdot \frac{d}{dx}(3 x^5 -6) = e^{3 x^5 -6} \cdot 15 x^4 \] Next, find \( v'(x) \): \[ v'(x) = \sec^2(x^2 + 9) \cdot \frac{d}{dx}(x^2 + 9) = \sec^2(x^2 + 9) \cdot 2x \] Now apply the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) = \left( e^{3 x^5 -6} \cdot 15 x^4 \right) \tan(x^2 + 9) + e^{3 x^5 -6} \cdot \left( \sec^2(x^2 + 9) \cdot 2x \right) \] Simplify and factor out the common term \( e^{3 x^5 -6} \): \[ f'(x) = e^{3 x^5 -6} \left( 15 x^4 \tan(x^2 + 9) + 2x \sec^2(x^2 + 9) \right) \] So, the derivative of the given function is: \[ \boxed{e^{3 x^5 -6}\left(15x^{4}\tan\left(x^{2}+9\right)+2x\sec^{2}\left(x^{2}+9\right)\right)} \] \(\boxed{e^{3 x^{5}-6}\left(15x^{4}\tan(x^{2}+9)+2x\,\sec^{2}(x^{2}+9)\right)}\) |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| kimi | 1. **Identify the function to be differentiated:** \[ f(x) = e^{3x^5 - 6} \cdot \tan(x^2 + 9) \] 2. **Apply the product rule for differentiation:** The product rule states that if \( u(x) \) and \( v(x) \) are differentiable functions, then: \[ (uv)' = u'v + uv' \] Here, let \( u(x) = e^{3x^5 - 6} \) and \( v(x) = \tan(x^2 + 9) \). 3. **Differentiate \( u(x) = e^{3x^5 - 6} \):** Using the chain rule, where \( u = e^w \) and \( w = 3x^5 - 6 \): \[ u'(x) = e^{3x^5 - 6} \cdot (3x^5 - 6)' = e^{3x^5 - 6} \cdot 15x^4 \] 4. **Differentiate \( v(x) = \tan(x^2 + 9) \):** Again, using the chain rule, where \( v = \tan(z) \) and \( z = x^2 + 9 \): \[ v'(x) = \sec^2(x^2 + 9) \cdot (x^2 + 9)' = \sec^2(x^2 + 9) \cdot 2x \] 5. **Apply the product rule:** \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substituting the derivatives: \[ f'(x) = \left( e^{3x^5 - 6} \cdot 15x^4 \right) \cdot \tan(x^2 + 9) + \left( e^{3x^5 - 6} \cdot \tan(x^2 + 9) \right) \cdot \left( \sec^2(x^2 + 9) \cdot 2x \right) \] 6. **Simplify the expression:** \[ f'(x) = 15x^4 e^{3x^5 - 6} \tan(x^2 + 9) + 2x e^{3x^5 - 6} \tan(x^2 + 9) \sec^2(x^2 + 9) \] 7. **Factor out common terms:** \[ f'(x) = e^{3x^5 - 6} \tan(x^2 + 9) \left( 15x^4 + 2x \sec^2(x^2 + 9) \right) \] 8. **Final answer:** \[ \boxed{f'(x) = e^{3x^5 - 6} \tan(x^2 + 9) \left( 15x^4 + 2x \sec^2(x^2 + 9) \right)} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| minimax | To differentiate the function $f(x) = e^{3x^5-6} \tan(x^2+9)$, 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 break down the function into two parts: $u(x) = e^{3x^5-6}$ and $v(x) = \tan(x^2+9)$. First, we find the derivatives of $u(x)$ and $v(x)$: 1. The derivative of $u(x)$ with respect to $x$ is: $$ u'(x) = \frac{d}{dx} e^{3x^5-6} = e^{3x^5-6} \cdot \frac{d}{dx}(3x^5-6) = e^{3x^5-6} \cdot (15x^4) $$ 2. The derivative of $v(x)$ with respect to $x$ is: $$ v'(x) = \frac{d}{dx} |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| wenxinyiyan | 首先，我们需要使用乘积法则来求这个函数的导数。乘积法则告诉我们如何求两个函数的乘积的导数。 假设我们有两个函数u(x)和v(x)，那么它们的乘积的导数就是： (u × v)' = u' × v + u × v' 在这个问题中，我们的u(x)是$e^{3 x^5-6}$，v(x)是$\tan \left(x^2+9\right)$。 首先，我们求$u(x) = e^{3 x^5-6}$的导数： $u'(x) = e^{3 x^5-6} × (3 x^5-6)' = e^{3 x^5-6} × 15x^4$ 接着，我们求$v(x) = \tan \left(x^2+9\right)$的导数： $v'(x) = \sec^2 \left(x^2+9\right) × (x^2+9)' = \sec^2 \left(x^2+9\right) × 2x$ 现在，我们可以使用乘积法则来找到整个函数的导数： $f'(x) = u'(x) × v(x) + u(x) × v'(x)$ $= e^{3 x^5-6} × 15x^4 × \tan \left(x^2+9\right) + e^{3 x^5-6} × \sec^2 \left(x^2+9\right) × 2x$ 所以，最终答案是： $\boxed{f'(x) = e^{3 x^5-6} × 15x^4 × \tan \left(x^2+9\right) + e^{3 x^5-6} × \sec^2 \left(x^2+9\right) × 2x}$ |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| zhipu | To differentiate the function \( f(x) = e^{3x^5 - 6} \tan(x^2 + 9) \), we will use the product rule and the chain rule. The product rule states that if you have a function \( u(x) \cdot v(x) \), its derivative is given by: \[ (uv)' = u'v + uv' \] Here, let \( u(x) = e^{3x^5 - 6} \) and \( v(x) = \tan(x^2 + 9) \). First, we find the derivatives of \( u(x) \) and \( v(x) \): 1. **Derivative of \( u(x) = e^{3x^5 - 6} \):** Using the chain rule, if \( u(x) = e^{g(x)} \), then \( u'(x) = e^{g(x)} \cdot g'(x) \). Here, \( g(x) = 3x^5 - 6 \), so: \[ g'(x) = 15x^4 \] Therefore, \[ u'(x) = e^{3x^5 - 6} \cdot 15x^4 \] 2. **Derivative of \( v(x) = \tan(x^2 + 9) \):** Using the chain rule again, if \( v(x) = \tan(h(x)) \), then \( v'(x) = \sec^2(h(x)) \cdot h'(x) \). Here, \( h(x) = x^2 + 9 \), so: \[ h'(x) = 2x \] Therefore, \[ v'(x) = \sec^2(x^2 + 9) \cdot 2x \] Now, applying the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substituting \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ f'(x) = \left( e^{3x^5 - 6} \cdot 15x^4 \right) \tan(x^2 + 9) + e^{3x^5 - 6} \left( \sec^2(x^2 + 9) \cdot 2x \right) \] Simplifying, we get: \[ f'(x) = e^{3x^5 - 6} \cdot 15x^4 \tan(x^2 + 9) + 2x e^{3x^5 - 6} \sec^2(x^2 + 9) \] Factoring out the common term \( e^{3x^5 - 6} \): \[ f'(x) = e^{3x^5 - 6} \left( 15x^4 \tan(x^2 + 9) + 2x \sec^2(x^2 + 9) \right) \] Thus, the derivative of the function is: \[ \boxed{e^{3x^5 - 6} \left( 15x^4 \tan(x^2 + 9) + 2x \sec^2(x^2 + 9) \right)} \] |
e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-0b53f04c-55da-9302-ab1d-7481fc0709d8', 'request_id': '0b53f04c-55da-9302-ab1d-7481fc0709d8'} | e^{3 x^5-6} x \left(2 \sec ^2\left(x^2+9\right)+15 x^3 \tan \left(x^2+9\right)\right) |