| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To differentiate the function \(\sqrt[3]{5-8x^2}\), we will use the chain rule. Let's proceed step by step. 1. **Rewrite the function in a more convenient form:** \[ \sqrt[3]{5-8x^2} = (5-8x^2)^{\frac{1}{3}} \] 2. **Apply the chain rule:** The chain rule states that if \(y = f(g(x))\), then \(y' = f'(g(x)) \cdot g'(x)\). Here, let: \[ u = 5 - 8x^2 \] Then: \[ y = u^{\frac{1}{3}} \] 3. **Differentiate \(y\) with respect to \(u\):** \[ \frac{dy}{du} = \frac{d}{du} \left( u^{\frac{1}{3}} \right) = \frac{1}{3} u^{-\frac{2}{3}} \] 4. **Differentiate \(u\) with respect to \(x\):** \[ \frac{du}{dx} = \frac{d}{dx} \left( 5 - 8x^2 \right) = -16x \] 5. **Combine the results using the chain rule:** \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \left( \frac{1}{3} u^{-\frac{2}{3}} \right) \cdot (-16x) \] 6. **Substitute \(u\) back into the expression:** \[ u = 5 - 8x^2 \] \[ \frac{dy}{dx} = \frac{1}{3} (5 - 8x^2)^{-\frac{2}{3}} \cdot (-16x) \] 7. **Simplify the expression:** \[ \frac{dy}{dx} = -\frac{16x}{3} (5 - 8x^2)^{-\frac{2}{3}} \] Thus, the derivative of the function \(\sqrt[3]{5-8x^2}\) is: \[ \boxed{-\frac{16x}{3} (5 - 8x^2)^{-\frac{2}{3}}} \] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| deepseek-r1 | To differentiate the function \(\sqrt[3]{5 - 8x^2}\), we start by rewriting it using exponents: \((5 - 8x^2)^{1/3}\). Using the chain rule, we first take the derivative of the outer function \(u^{1/3}\) where \(u = 5 - 8x^2\). The derivative of \(u^{1/3}\) is \(\frac{1}{3}u^{-2/3}\). Next, we multiply by the derivative of the inner function \(5 - 8x^2\), which is \(-16x\). Putting it all together, we get: \[ \frac{d}{dx} \left( (5 - 8x^2)^{1/3} \right) = \frac{1}{3}(5 - 8x^2)^{-2/3} \cdot (-16x) \] Simplifying the expression, we combine the coefficients: \[ \frac{1}{3} \cdot (-16x) = -\frac{16x}{3} \] Thus, the derivative is: \[ -\frac{16x}{3}(5 - 8x^2)^{-2/3} \] Rewriting this in a more standard form, we have: \[ \boxed{-\dfrac{16x}{3(5 - 8x^2)^{2/3}}} \] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| gpto1 | To find the derivative of the function \( y = \sqrt[3]{5 - 8x^2} \), let's apply the chain rule. We can rewrite the function as \( y = (5 - 8x^2)^{1/3} \). **Step 1: Differentiate the outer function** The outer function is \( u^{1/3} \), where \( u = 5 - 8x^2 \). The derivative of the outer function with respect to \( u \) is: \[ \frac{d}{du} u^{1/3} = \frac{1}{3} u^{-2/3} \] **Step 2: Differentiate the inner function** The inner function is \( u = 5 - 8x^2 \). The derivative of the inner function with respect to \( x \) is: \[ \frac{du}{dx} = -16x \] **Step 3: Apply the chain rule** Multiply the derivative of the outer function by the derivative of the inner function: \[ \frac{dy}{dx} = \frac{1}{3} (5 - 8x^2)^{-2/3} \cdot (-16x) \] Simplify the constants: \[ \frac{1}{3} \cdot (-16x) = -\frac{16x}{3} \] So the derivative is: \[ \frac{dy}{dx} = -\frac{16x}{3} (5 - 8x^2)^{-2/3} \] This can also be written as: \[ \frac{dy}{dx} = \frac{-16x}{3(5 - 8x^2)^{2/3}} \] **Final Answer:** \(\boxed{\dfrac{-16x}{3(5\,-\,8x^{2})^{2/3}}}\) |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| gpt-4o-2024-08-06 | To differentiate the function \( y = \sqrt[3]{5 - 8x^2} \), we first rewrite it in a form that is easier to differentiate using the power rule. The cube root can be expressed as a power of \( \frac{1}{3} \): \[ y = (5 - 8x^2)^{\frac{1}{3}} \] Now, we apply the chain rule to differentiate this function. The chain rule states that if \( y = f(g(x)) \), then the derivative \( y' \) is given by: \[ y' = f'(g(x)) \cdot g'(x) \] In our case, let \( u = 5 - 8x^2 \), so that \( y = u^{\frac{1}{3}} \). We need to find \( \frac{dy}{dx} \), which involves finding \( \frac{dy}{du} \) and \( \frac{du}{dx} \). First, differentiate \( y = u^{\frac{1}{3}} \) with respect to \( u \): \[ \frac{dy}{du} = \frac{1}{3} u^{-\frac{2}{3}} \] Next, differentiate \( u = 5 - 8x^2 \) with respect to \( x \): \[ \frac{du}{dx} = -16x \] Now, apply the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{3} u^{-\frac{2}{3}} \cdot (-16x) \] Substitute back \( u = 5 - 8x^2 \): \[ \frac{dy}{dx} = \frac{1}{3} (5 - 8x^2)^{-\frac{2}{3}} \cdot (-16x) \] Simplify the expression: \[ \frac{dy}{dx} = -\frac{16x}{3} (5 - 8x^2)^{-\frac{2}{3}} \] Thus, the derivative of the function is: \[ \boxed{-\frac{16x}{3} (5 - 8x^2)^{-\frac{2}{3}}} \] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| DeepSeek-Coder-V2-Lite-Instruct | The differentiation of the function \(\sqrt[3]{5-8 x^2}\) can be found using the chain rule. Let's denote the function inside the cube root as \(u = 5 - 8x^2\). The derivative of \(\sqrt[3]{u}\) with respect to \(u\) is \(\frac{1}{3u^{2/3}}\). Therefore, we need to apply the chain rule to find the derivative with respect to \(x\). First, we find the derivative of \(u\) with respect to \(x\): \[ \frac{du}{dx} = \frac{d}{dx}(5 - 8x^2) = -16x \] Next, we use the chain rule to find the derivative of \(\sqrt[3]{u}\) with respect to \(x\): \[ \frac{d}{dx} \left( \sqrt[3]{u} \right) = \frac{1}{3u^{2/3}} \cdot \frac{du}{dx} \] Substituting \(u = 5 - 8x^2\) and \(\frac{du}{dx} = -16x\) into the equation, we get: \[ \frac{d}{dx} \left( \sqrt[3]{5-8x^2} \right) = \frac{1}{3(5-8x^2)^{2/3}} \cdot (-16x) \] Simplifying this expression, we obtain: \[ \frac{d}{dx} \left( \sqrt[3]{5-8x^2} \right) = -\frac{16x}{3(5-8x^2)^{2/3}} \] Thus, the final answer is: \[ \boxed{-\frac{16x}{3(5-8x^2)^{2/3}}} \] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The function given is $\sqrt[3]{5-8 x^2}$. ## Step 2: Apply the chain rule for differentiation To differentiate this function, we will apply the chain rule. The chain rule states that if we have a composite function $f(g(x))$, then its derivative is $f'(g(x)) \cdot g'(x)$. ## Step 3: Identify the outer and inner functions The outer function is $\sqrt[3]{u}$, where $u = 5-8 x^2$. The inner function is $u = 5-8 x^2$. ## Step 4: Differentiate the outer function The derivative of the outer function $\sqrt[3]{u}$ is $\frac{1}{3}u^{-\frac{2}{3}}$. ## Step 5: Differentiate the inner function The derivative of the inner function $u = 5-8 x^2$ is $u' = -16x$. ## Step 6: Apply the chain rule Using the chain rule, the derivative of the given function is $\frac{1}{3}u^{-\frac{2}{3}} \cdot u'$. ## Step 7: Substitute the expressions for $u$ and $u'$ Substituting $u = 5-8 x^2$ and $u' = -16x$, we get $\frac{1}{3}(5-8 x^2)^{-\frac{2}{3}} \cdot (-16x)$. ## Step 8: Simplify the expression Simplifying the expression, we get $\frac{-16x}{3(5-8 x^2)^{\frac{2}{3}}}$. ## Step 9: Write the final answer in the required format The final answer is: $\boxed{\frac{-16x}{3(5-8 x^2)^{\frac{2}{3}}}}$ |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the function to be differentiated The given function is $\sqrt[3]{5-8 x^2}$. ## Step 2: Apply the chain rule for differentiation To differentiate the given function, we will apply the chain rule. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative is given by $f'(g(x)) \cdot g'(x)$. ## Step 3: Identify the outer and inner functions In the given function, the outer function is $\sqrt[3]{u}$ and the inner function is $u = 5-8 x^2$. ## Step 4: Differentiate the outer function The derivative of the outer function $\sqrt[3]{u}$ with respect to $u$ is $\frac{1}{3}u^{-\frac{2}{3}}$. ## Step 5: Differentiate the inner function The derivative of the inner function $u = 5-8 x^2$ with respect to $x$ is $-16x$. ## Step 6: Apply the chain rule Using the chain rule, the derivative of the given function is $\frac{1}{3}u^{-\frac{2}{3}} \cdot (-16x)$. ## Step 7: Substitute the inner function back into the derivative Substituting $u = 5-8 x^2$ back into the derivative, we get $\frac{1}{3}(5-8 x^2)^{-\frac{2}{3}} \cdot (-16x)$. ## Step 8: Simplify the derivative Simplifying the derivative, we get $\frac{-16x}{3(5-8 x^2)^{\frac{2}{3}}}$. ## Step 9: Write the final answer in the required format The final answer is: $\boxed{\frac{-16x}{3(5-8 x^2)^{\frac{2}{3}}}}$ |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| Qwen2-72B-Instruct | To differentiate the function \(f(x) = \sqrt[3]{5-8x^2}\), we can use the chain rule. The function can be rewritten as \(f(x) = (5-8x^2)^{\frac{1}{3}}\). Let's denote \(u = 5-8x^2\). Then, \(f(u) = u^{\frac{1}{3}}\). First, we find the derivative of \(f(u)\) with respect to \(u\): \[f'(u) = \frac{d}{du}(u^{\frac{1}{3}}) = \frac{1}{3}u^{-\frac{2}{3}}\] Next, we find the derivative of \(u\) with respect to \(x\): \[\frac{du}{dx} = \frac{d}{dx}(5-8x^2) = -16x\] Now, using the chain rule, we have: \[f'(x) = f'(u) \cdot \frac{du}{dx} = \left(\frac{1}{3}u^{-\frac{2}{3}}\right) \cdot (-16x)\] Substituting back \(u = 5-8x^2\), we get: \[f'(x) = \left(\frac{1}{3}(5-8x^2)^{-\frac{2}{3}}\right) \cdot (-16x)\] Simplifying further: \[f'(x) = -\frac{16x}{3(5-8x^2)^{\frac{2}{3}}}\] Therefore, the derivative of the function is: \[f'(x) = \boxed{-\frac{16x}{3(5-8x^2)^{\frac{2}{3}}}}\] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| baichuan4 | To differentiate the function $f(x) = \sqrt[3]{5-8x^2}$, we can use the chain rule. The chain rule states that if we have a function $f(g(x))$, then its derivative is $f'(g(x)) \cdot g'(x)$. First, let's identify our inner and outer functions. The outer function is $f(u) = u^{1/3}$, and the inner function is $g(x) = 5 - 8x^2$. We need to find the derivatives of both functions. The derivative of the outer function $f(u) = u^{1/3}$ is $f'(u) = \frac{1}{3}u^{-2/3}$. The derivative of the inner function $g(x) = 5 - 8x^2$ is $g'(x) = -16x$. Now, we apply the chain rule to find the derivative of the original function: $$ \frac{d}{dx}\left[\sqrt[3]{5-8x^2}\right] = \frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x). $$ Substituting the derivatives we found earlier, we get: $$ \frac{d}{dx}\left[\sqrt[3]{5-8x^2}\right] = \frac{1}{3}(5-8x^2)^{-2/3} \cdot (-16x). $$ Simplifying this expression, we have: $$ \frac{d}{dx}\left[\sqrt[3]{5-8x^2}\right] = -\frac{16x}{3(5-8x^2)^{2/3}}. $$ This is the derivative of the function $\sqrt[3]{5-8x^2}$. To put the final answer in a box as requested, we have: $$ \boxed{-\frac{16x}{3(5-8x^2)^{2/3}}}. $$ |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| doubao | Let $y=\sqrt[3]{5-8x^2}=(5-8x^2)^{\frac{1}{3}}$. Using the chain rule, let $u=5-8x^2$, then $y=u^{\frac{1}{3}}$. The derivative of $u$ with respect to $x$ is $u'=-16x$. The derivative of $y$ with respect to $u$ is $y'=\frac{1}{3}u^{-\frac{2}{3}}$. So the derivative of $y$ with respect to $x$ is: \[ \begin{align*} y'&=\frac{1}{3}(5-8x^2)^{-\frac{2}{3}}\times(-16x)\\ &=\frac{-16x}{3(5-8x^2)^{\frac{2}{3}}}\\ &=\boxed{\frac{-16x}{3(5-8x^2)^{\frac{2}{3}}}} \end{align*} \] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| gpto1 | To find the derivative of the function \( y = \sqrt[3]{5 - 8x^2} \), let's apply the chain rule. We can rewrite the function as \( y = (5 - 8x^2)^{1/3} \). **Step 1: Differentiate the outer function** The outer function is \( u^{1/3} \), where \( u = 5 - 8x^2 \). The derivative of the outer function with respect to \( u \) is: \[ \frac{d}{du} u^{1/3} = \frac{1}{3} u^{-2/3} \] **Step 2: Differentiate the inner function** The inner function is \( u = 5 - 8x^2 \). The derivative of the inner function with respect to \( x \) is: \[ \frac{du}{dx} = -16x \] **Step 3: Apply the chain rule** Multiply the derivative of the outer function by the derivative of the inner function: \[ \frac{dy}{dx} = \frac{1}{3} (5 - 8x^2)^{-2/3} \cdot (-16x) \] Simplify the constants: \[ \frac{1}{3} \cdot (-16x) = -\frac{16x}{3} \] So the derivative is: \[ \frac{dy}{dx} = -\frac{16x}{3} (5 - 8x^2)^{-2/3} \] This can also be written as: \[ \frac{dy}{dx} = \frac{-16x}{3(5 - 8x^2)^{2/3}} \] **Final Answer:** \(\boxed{\dfrac{-16x}{3(5\,-\,8x^{2})^{2/3}}}\) |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| kimi | To differentiate the function \( f(x) = \sqrt[3]{5 - 8x^2} \), we will use the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative \( f'(x) \) is given by: \[ f'(x) = f'(g(x)) \cdot g'(x) \] Here, our function \( f(x) \) can be written as: \[ f(x) = (5 - 8x^2)^{1/3} \] Let's identify the inner function \( g(x) \) and the outer function \( h(u) \): \[ g(x) = 5 - 8x^2 \] \[ h(u) = u^{1/3} \] First, we differentiate the outer function \( h(u) \) with respect to \( u \): \[ h'(u) = \frac{d}{du} (u^{1/3}) = \frac{1}{3} u^{-2/3} \] Next, we differentiate the inner function \( g(x) \) with respect to \( x \): \[ g'(x) = \frac{d}{dx} (5 - 8x^2) = -16x \] Now, we apply the chain rule: \[ f'(x) = h'(g(x)) \cdot g'(x) \] Substitute \( g(x) = 5 - 8x^2 \) into \( h'(u) \): \[ h'(g(x)) = \frac{1}{3} (5 - 8x^2)^{-2/3} \] Then, multiply by \( g'(x) \): \[ f'(x) = \frac{1}{3} (5 - 8x^2)^{-2/3} \cdot (-16x) \] Simplify the expression: \[ f'(x) = -\frac{16x}{3} (5 - 8x^2)^{-2/3} \] To make it more readable, we can write the negative exponent as a fraction: \[ f'(x) = -\frac{16x}{3 (5 - 8x^2)^{2/3}} \] Thus, the final answer is: \[ \boxed{-\frac{16x}{3 (5 - 8x^2)^{2/3}}} \] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| minimax | To differentiate the function $f(x) = \sqrt[3]{5-8x^2}$, we can rewrite it using the cube root function and then apply the chain rule. First, let's rewrite the function as $f(x) = (5-8x^2)^{\frac{1}{3}}$. Now, we'll differentiate it using the chain rule. The chain rule states that if you have a composite function $f(g(x))$, then its derivative is $f'(g(x)) \cdot g'(x)$. Let $u = 5 - 8x^2$ and $f(u) = u^{\frac{1}{3}}$. First, we find the derivative of $u$ with respect to $x$: $$\frac{du}{dx} = -16x.$$ Next, we find the derivative of $f(u)$ with respect to $u$: $$\frac{df}{du} = \frac{1}{3}u^{-\frac{2}{3}}.$$ Now, we apply the chain rule: $$\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}.$$ Sub |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| wenxinyiyan | 首先，将函数写为$(5-8 x^2)^{\frac{1}{3}}$。 然后，使用链式法则进行求导。 链式法则公式为：$\frac{d}{dx}f(g(x))=f^{\prime}(g(x))g^{\prime}(x)$。 在这里，$f(u)=u^{\frac{1}{3}}$，$g(x)=5-8 x^2$。 首先求$f^{\prime}(u)$和$g^{\prime}(x)$： $f^{\prime}(u)=\frac{1}{3}u^{-\frac{2}{3}}$ $g^{\prime}(x)=-16x$ 然后，应用链式法则： $\frac{d}{dx}(5-8 x^2)^{\frac{1}{3}}=\frac{1}{3}(5-8 x^2)^{-\frac{2}{3}}(-16x)$ 化简得： $\boxed{\frac{-16x}{3\sqrt[3]{(5-8 x^2)^2}}}$ |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| zhipu | To differentiate the function \( f(x) = \sqrt[3]{5 - 8x^2} \), we can use the chain rule. Let's break it down step by step. 1. Rewrite the function in a more convenient form for differentiation: \[ f(x) = (5 - 8x^2)^{1/3} \] 2. Apply the chain rule. The chain rule states that if you have a function \( g(h(x)) \), its derivative is \( g'(h(x)) \cdot h'(x) \). Here, let: \[ u = 5 - 8x^2 \quad \text{so that} \quad f(x) = u^{1/3} \] 3. Differentiate \( u^{1/3} \) with respect to \( u \): \[ \frac{d}{du}(u^{1/3}) = \frac{1}{3}u^{-2/3} \] 4. Differentiate \( u = 5 - 8x^2 \) with respect to \( x \): \[ \frac{d}{dx}(5 - 8x^2) = -16x \] 5. Combine these results using the chain rule: \[ f'(x) = \frac{d}{du}(u^{1/3}) \cdot \frac{du}{dx} = \frac{1}{3}u^{-2/3} \cdot (-16x) \] 6. Substitute back \( u = 5 - 8x^2 \): \[ f'(x) = \frac{1}{3}(5 - 8x^2)^{-2/3} \cdot (-16x) \] 7. Simplify the expression: \[ f'(x) = -\frac{16x}{3}(5 - 8x^2)^{-2/3} \] Therefore, the derivative of the function \( \sqrt[3]{5 - 8x^2} \) is: \[ \boxed{-\frac{16x}{3}(5 - 8x^2)^{-2/3}} \] |
-\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-9b47ce1c-5f1c-947f-bb44-71957a7f8270', 'request_id': '9b47ce1c-5f1c-947f-bb44-71957a7f8270'} | -\frac{16 x}{3 \sqrt[3]{5-8 x^2}^2} |