| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the derivative of the function \( g(x) = \frac{1}{1 + x^{3/2}} \) with respect to \( x \), we will apply the chain rule and the quotient rule. Given: \[ g(x) = \frac{1}{1 + x^{3/2}} = (1 + x^{3/2})^{-1} \] Let's differentiate \( g(x) \) with respect to \( x \): \[ g'(x) = -1 \cdot (1 + x^{3/2})^{-2} \cdot \frac{3}{2}x^{1/2} \] \[ g'(x) = -\frac{3}{2} \cdot \frac{x^{1/2}}{(1 + x^{3/2})^2} \] Now, let's find \( g'(1) \): \[ g'(1) = -\frac{3}{2} \cdot \frac{1^{1/2}}{(1 + 1^{3/2})^2} \] \[ g'(1) = -\frac{3}{2} \cdot \frac{1}{(1 + 1)^2} \] \[ g'(1) = -\frac{3}{2} \cdot \frac{1}{4} \] \[ g'(1) = -\frac{3}{8} \] Therefore, the derivative of \( g(x) \) at \( x = 1 \) is \( -\frac{3}{8} \). |
-0.375 |
| deepseekr1 | To find the derivative of the function \( g(x) = \frac{1}{1 + x^{3/2}} \) at \( x = 1 \), we can use the quotient rule or the chain rule. Here, we use the quotient rule: 1. **Quotient Rule Application**: - The function \( g(x) \) is written as \( \frac{1}{1 + x^{3/2}} \). - Let \( f(x) = 1 \) and \( h(x) = 1 + x^{3/2} \). - The derivatives are \( f'(x) = 0 \) and \( h'(x) = \frac{3}{2}x^{1/2} \). 2. **Applying the Quotient Rule Formula**: \[ g'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^2} = \frac{0 \cdot (1 + x^{3/2}) - 1 \cdot \frac{3}{2}x^{1/2}}{(1 + x^{3/2})^2} = \frac{-\frac{3}{2}x^{1/2}}{(1 + x^{3/2})^2} \] 3. **Evaluating at \( x = 1 \)**: - Substitute \( x = 1 \) into the derivative: \[ g'(1) = \frac{-\frac{3}{2} \cdot 1^{1/2}}{(1 + 1^{3/2})^2} = \frac{-\frac{3}{2}}{(1 + 1)^2} = \frac{-\frac{3}{2}}{4} = -\frac{3}{8} \] 4. **Verification with Numerical Approximation**: - Using the central difference formula with a small \( h = 0.001 \), the numerical derivative approximates to -0.374, which is very close to the exact value -0.375 (or -3/8). Thus, the value of \( g'(1) \) is \(\boxed{-\dfrac{3}{8}}\). |
-0.375 |
| gpto1 | To find \( g'(x) \) when \( x = 1 \) for the function \( g(x) = \dfrac{1}{1 + x^{3/2}} \), we will follow these steps: **Step 1: Rewrite the Function** First, express \( g(x) \) in a form that makes differentiation straightforward: \[ g(x) = [1 + x^{3/2}]^{-1} \] **Step 2: Differentiate Using the Chain Rule** Use the chain rule to find \( g'(x) \): \[ g'(x) = -[1 + x^{3/2}]^{-2} \cdot \dfrac{d}{dx}[1 + x^{3/2}] \] Compute the derivative of the inner function \( u(x) = 1 + x^{3/2} \): \[ u'(x) = \dfrac{d}{dx}[1 + x^{3/2}] = \dfrac{d}{dx}[x^{3/2}] = \dfrac{3}{2} x^{1/2} \] So the derivative is: \[ g'(x) = -[1 + x^{3/2}]^{-2} \cdot \dfrac{3}{2} x^{1/2} \] **Step 3: Simplify and Evaluate at \( x = 1 \)** Plug in \( x = 1 \): - Compute \( x^{1/2} \): \[ x^{1/2} = 1^{1/2} = 1 \] - Compute \( x^{3/2} \): \[ x^{3/2} = 1^{3/2} = 1 \] - Compute \( [1 + x^{3/2}]^{-2} \): \[ [1 + 1]^{-2} = [2]^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4} \] Now, substitute these values into \( g'(x) \): \[ g'(1) = -\left( \dfrac{3}{2} \times 1 \right) \times \dfrac{1}{4} = -\dfrac{3}{2} \times \dfrac{1}{4} \] Simplify the expression: \[ g'(1) = -\dfrac{3}{2} \times \dfrac{1}{4} = -\dfrac{3}{8} \] **Answer:** \[ g'(1) = -\dfrac{3}{8} \] |
-0.375 |
| GPT-4o | To find the derivative of the function \( g(x) = \frac{1}{1 + x^{3/2}} \), we can use the chain rule and the quotient rule. Here's the step-by-step solution: First, rewrite the function in a form that is easier to differentiate using the chain rule: \[ g(x) = (1 + x^{3/2})^{-1} \] Now, let's differentiate \( g(x) \) with respect to \( x \): 1. **Apply the chain rule**: \[ g'(x) = -1 \cdot (1 + x^{3/2})^{-2} \cdot \frac{d}{dx}(1 + x^{3/2}) \] 2. **Differentiate the inner function \( 1 + x^{3/2} \)**: \[ \frac{d}{dx}(1 + x^{3/2}) = \frac{d}{dx}(1) + \frac{d}{dx}(x^{3/2}) \] The derivative of 1 is 0, and the derivative of \( x^{3/2} \) using the power rule is: \[ \frac{d}{dx}(x^{3/2}) = \frac{3}{2} x^{1/2} \] 3. **Substitute back into the chain rule expression**: \[ g'(x) = - (1 + x^{3/2})^{-2} \cdot \frac{3}{2} x^{1/2} \] \[ g'(x) = - \frac{3}{2} x^{1/2} (1 + x^{3/2})^{-2} \] Now, we need to evaluate \( g'(x) \) at \( x = 1 \): 4. **Substitute \( x = 1 \) into \( g'(x) \)**: \[ g'(1) = - \frac{3}{2} \cdot 1^{1/2} \cdot (1 + 1^{3/2})^{-2} \] \[ g'(1) = - \frac{3}{2} \cdot 1 \cdot (1 + 1)^{-2} \] \[ g'(1) = - \frac{3}{2} \cdot (2)^{-2} \] \[ g'(1) = - \frac{3}{2} \cdot \frac{1}{4} \] \[ g'(1) = - \frac{3}{8} \] So, the value of the derivative \( g'(x) \) at \( x = 1 \) is: \[ g'(1) = - \frac{3}{8} \] |
-0.375 |
| llama3.1-70B-Instruct | To find the derivative of g(x) = 1 / (1 + x^{3/2}) and evaluate it at x = 1, we'll apply the quotient rule and chain rule of differentiation. The quotient rule states that if we have a function of the form: f(x) = u(x) / v(x) Then its derivative f'(x) is given by: f'(x) = (u'(x)v(x) - u(x)v'(x)) / v(x)^2 In our case, u(x) = 1 and v(x) = 1 + x^{3/2}. We need to find u'(x) and v'(x). u'(x) = d(1)/dx = 0 (since the derivative of a constant is 0) v'(x) = d(1 + x^{3/2})/dx = (3/2)x^{1/2} (using the power rule for differentiation) Now, we can apply the quotient rule: g'(x) = (u'(x)v(x) - u(x)v'(x)) / v(x)^2 = (0 * (1 + x^{3/2}) - 1 * (3/2)x^{1/2}) / (1 + x^{3/2})^2 = -(3/2)x^{1/2} / (1 + x^{3/2})^2 To find g'(1), we substitute x = 1 into the expression for g'(x): g'(1) = -(3/2)(1)^{1/2} / (1 + (1)^{3/2})^2 = -(3/2) / (1 + 1)^2 = -(3/2) / 4 = -3/8 Therefore, g'(1) = -3/8. |
-0.375 |
| llama3.1-405B-Instruct | To find the derivative of g(x) = 1 / (1 + x^{3/2}), we can use the quotient rule and the chain rule. First, let's rewrite g(x) as: g(x) = (1 + x^{3/2})^{-1} Now, we can use the chain rule to find the derivative of g(x): g'(x) = -1(1 + x^{3/2})^{-2} \* (3/2)x^{1/2} Simplifying, we get: g'(x) = -3/2 \* x^{1/2} / (1 + x^{3/2})^2 Now, we need to find g'(1). Plugging in x = 1, we get: g'(1) = -3/2 \* 1^{1/2} / (1 + 1^{3/2})^2 g'(1) = -3/2 / (1 + 1)^2 g'(1) = -3/2 / 4 g'(1) = -3/8 Therefore, g'(1) = -3/8. |
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| DeepSeek-Coder-V2 | To find the derivative of the function g(x) = 1 / (1 + x^{3/2}) at x = 1, we will use the quotient rule for differentiation. The quotient rule states that if we have a function h(x) = f(x) / g(x), then the derivative h'(x) is given by: h'(x) = (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2 In our case, f(x) = 1 and g(x) = 1 + x^{3/2}. Now, let's find the derivatives of f(x) and g(x): 1. f(x) = 1, so f'(x) = 0 2. g(x) = 1 + x^{3/2}, so g'(x) = (3/2) * x^{1/2} Now, we can apply the quotient rule to find g'(x): g'(x) = (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2 g'(x) = (0 * (1 + x^{3/2}) - 1 * (3/2) * x^{1/2}) / (1 + x^{3/2})^2 g'(x) = (-(3/2) * x^{1/2}) / (1 + x^{3/2})^2 Now, we need to evaluate g'(x) at x = 1: g'(1) = (-(3/2) * 1^{1/2}) / (1 + 1^{3/2})^2 g'(1) = (-(3/2) * 1) / (1 + 1)^2 g'(1) = (-(3/2)) / 2^2 g'(1) = (-(3/2)) / 4 g'(1) = -3/8 So, the value of g'(x) when x = 1 is -3/8. |
-0.375 |
| Qwen2-72B-Instruct | To find the derivative of the function \( g(x) = \frac{1}{1 + x^{3/2}} \) with respect to \( x \), we will apply the chain rule and the quotient rule. Given: \[ g(x) = \frac{1}{1 + x^{3/2}} = (1 + x^{3/2})^{-1} \] Let's differentiate \( g(x) \) with respect to \( x \): \[ g'(x) = -1 \cdot (1 + x^{3/2})^{-2} \cdot \frac{3}{2}x^{1/2} \] \[ g'(x) = -\frac{3}{2} \cdot \frac{x^{1/2}}{(1 + x^{3/2})^2} \] Now, let's find \( g'(1) \): \[ g'(1) = -\frac{3}{2} \cdot \frac{1^{1/2}}{(1 + 1^{3/2})^2} \] \[ g'(1) = -\frac{3}{2} \cdot \frac{1}{(1 + 1)^2} \] \[ g'(1) = -\frac{3}{2} \cdot \frac{1}{4} \] \[ g'(1) = -\frac{3}{8} \] Therefore, the derivative of \( g(x) \) at \( x = 1 \) is \( -\frac{3}{8} \). |
-0.375 |
| baichuan4 | To find the derivative \(g'(x)\) of the function \(g(x) = \frac{1}{1 + x^{\frac{3}{2}}}\), we can use the chain rule. The function \(g(x)\) can be seen as a composition of two functions: 1. \(u = 1 + x^{\frac{3}{2}}\) 2. \(v = \frac{1}{u}\) We need to find the derivatives of these two functions and multiply them: 1. \(u' = \frac{3}{2}x^{-\frac{1}{2}}\) 2. \(v' = -\frac{1}{u^2}\) Now, applying the chain rule: \[g'(x) = v' \cdot u' = -\frac{1}{u^2} \cdot \frac{3}{2}x^{-\frac{1}{2}} = -\frac{3}{2}x^{-\frac{1}{2}} \cdot \frac{1}{(1 + x^{\frac{3}{2}})^2}\] Simplifying: \[g'(x) = -\frac{3}{2} \cdot \frac{x^{-\frac{1}{2}}}{(1 + x^{\frac{3}{2}})^2}\] Now, to find \(g'(1)\), we substitute \(x = 1\): \[g'(1) = -\frac{3}{2} \cdot \frac{1^{-\frac{1}{2}}}{(1 + 1^{\frac{3}{2}})^2} = -\frac{3}{2} \cdot \frac{1}{(1 + 1)^2} = -\frac{3}{2} \cdot \frac{1}{4} = -\frac{3}{8}\] So, \(g'(1) = -\frac{3}{8}\). |
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| doubao | First find the derivative of g(x) using the chain rule. Let u = 1 + x^(3/2), then g(x) = 1/u. The derivative of u with respect to x is (3/2)x^(1/2). The derivative of g(x) is g'(x) = -1/u^2 * (3/2)x^(1/2). Substituting u = 1 + x^(3/2) gives g'(x) = -1/(1 + x^(3/2))^2 * (3/2)x^(1/2). When x = 1, g'(1) = -1/(1 + 1)^2 * (3/2)*1^(1/2) = -3/8. | -0.375 |
| minimax | To find the derivative of the function \( g(x) = \frac{1}{1 + x^{3/2}} \), we can use the chain rule and the quotient rule. The chain rule is used to differentiate composite functions, and the quotient rule is used to differentiate functions that are the quotient of two other functions. Let's first apply the chain rule to differentiate \( x^{3/2} \): \[ \frac{d}{dx} x^{3/2} = \frac{3}{2} x^{(3/2) - 1} = \frac{3}{2} x^{1/2} \] Now, let's apply the quotient rule to differentiate \( g(x) \). The quotient rule states that if you have a function \( h(x) = \frac{f(x)}{g(x)} \), then its derivative \( h'(x) \) is given by: \[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \] In our case, \( f(x) = 1 \) and \( g(x) = 1 + x^{3/2} \). The derivative of \( f(x) \) with respect to \( x \) is \( f'(x) = 0 \), and the derivative of \( g(x) \) with respect to \( x \) is \( g'(x) = \frac{3}{2} x^{1/2} \) (from the chain rule application above). Applying the quotient rule, we get: \[ g'(x) = \frac{0 \cdot (1 + x^{3/2}) - 1 \cdot \frac{3}{2} x^{1/2}}{(1 + x^{3/2})^2} \] Simplify the expression: \[ g'(x) = \frac{-\frac{3}{2} x^{1/2}}{(1 + x^{3/2})^2} \] Now we need to find \( g'(1) \): \[ g'(1) = \frac{-\frac{3}{2} \cdot 1^{1/2}}{(1 + 1^{3/2})^2} \] Since \( 1^{1/2} = 1 \) and \( 1^{3/2} = 1 \), we have: \[ g'(1) = \frac{-\frac{3}{2}}{(1 + 1)^2} \] \[ g'(1) = \frac{-\frac{3}{2}}{2^2} \] \[ g'(1) = \frac{-\frac{3}{2}}{4} \] \[ g'(1) = -\frac{3}{8} \] Therefore, the derivative of \( g(x) \) when \( x = 1 \) is \( g'(1) = -\frac{3}{8} \). |
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| kimi | To find the derivative of the function g(x) = 1 / (1 + x^(3/2)), we'll use the chain rule. The chain rule states that if we have a composite function, say h(x) = f(g(x)), then the derivative h'(x) is the derivative of f evaluated at g(x) times the derivative of g evaluated at x, which is written as: h'(x) = f'(g(x)) * g'(x) In this case, let's consider f(u) = 1/u and g(x) = 1 + x^(3/2). Then g(x) is our inner function and f(u) is the outer function. First, we find the derivative of f(u) with respect to u: f'(u) = d(1/u)/du = -1/u^2 Next, we find the derivative of g(x) with respect to x: g'(x) = d(1 + x^(3/2))/dx = (3/2) * x^(1/2) Now, we apply the chain rule: g'(x) = f'(g(x)) * g'(x) g'(x) = -1/(1 + x^(3/2))^2 * (3/2) * x^(1/2) Now, we substitute x = 1 to find g'(1): g'(1) = -1/(1 + 1^(3/2))^2 * (3/2) * 1^(1/2) g'(1) = -1/(1 + 1)^2 * (3/2) g'(1) = -1/(2)^2 * (3/2) g'(1) = -1/4 * (3/2) g'(1) = -3/8 So, the derivative of g(x) at x = 1 is -3/8. |
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| wenxinyiyan | 首先，我们需要找到函数g(x)的导数。函数g(x)是(1 + x^{3/2})的倒数，我们可以使用链式法则和幂规则来找到它的导数。 g'(x) = -3/2 * x^{1/2} / (1 + x^{3/2})^2 然后，我们将x = 1代入g'(x)中，得到： g'(1) = -3/2 * 1^{1/2} / (1 + 1^{3/2})^2 = -3/2 * 1 / (1 + 1)^2 = -3/2 * 1 / 2^2 = -3/2 * 1 / 4 = -3/8 所以，当x = 1时，g'(x)的值为-3/8。 |
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| zhipu | To find the derivative of the function \( g(x) = \frac{1}{1 + x^{3/2}} \), we can use the chain rule and the quotient rule. The quotient rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), its derivative is given by: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] In our case, \( u(x) = 1 \) and \( v(x) = 1 + x^{3/2} \). The derivative of \( u(x) \) with respect to \( x \) is \( u'(x) = 0 \), and the derivative of \( v(x) \) with respect to \( x \) can be found using the chain rule: \[ v'(x) = \frac{d}{dx}(1 + x^{3/2}) = 0 + \frac{3}{2}x^{1/2} \] Now we can apply the quotient rule: \[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] \[ g'(x) = \frac{0 \cdot (1 + x^{3/2}) - 1 \cdot \left(\frac{3}{2}x^{1/2}\right)}{(1 + x^{3/2})^2} \] \[ g'(x) = \frac{-\frac{3}{2}x^{1/2}}{(1 + x^{3/2})^2} \] Now we need to find the value of \( g'(x) \) when \( x = 1 \): \[ g'(1) = \frac{-\frac{3}{2}(1)^{1/2}}{(1 + (1)^{3/2})^2} \] \[ g'(1) = \frac{-\frac{3}{2} \cdot 1}{(1 + 1)^2} \] \[ g'(1) = \frac{-\frac{3}{2}}{2^2} \] \[ g'(1) = \frac{-\frac{3}{2}}{4} \] \[ g'(1) = -\frac{3}{8} \] So the value of \( g'(x) \) when \( x = 1 \) is \( -\frac{3}{8} \). |
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| gpto1 | To find \( g'(x) \) when \( x = 1 \) for the function \( g(x) = \dfrac{1}{1 + x^{3/2}} \), we will follow these steps: **Step 1: Rewrite the Function** First, express \( g(x) \) in a form that makes differentiation straightforward: \[ g(x) = [1 + x^{3/2}]^{-1} \] **Step 2: Differentiate Using the Chain Rule** Use the chain rule to find \( g'(x) \): \[ g'(x) = -[1 + x^{3/2}]^{-2} \cdot \dfrac{d}{dx}[1 + x^{3/2}] \] Compute the derivative of the inner function \( u(x) = 1 + x^{3/2} \): \[ u'(x) = \dfrac{d}{dx}[1 + x^{3/2}] = \dfrac{d}{dx}[x^{3/2}] = \dfrac{3}{2} x^{1/2} \] So the derivative is: \[ g'(x) = -[1 + x^{3/2}]^{-2} \cdot \dfrac{3}{2} x^{1/2} \] **Step 3: Simplify and Evaluate at \( x = 1 \)** Plug in \( x = 1 \): - Compute \( x^{1/2} \): \[ x^{1/2} = 1^{1/2} = 1 \] - Compute \( x^{3/2} \): \[ x^{3/2} = 1^{3/2} = 1 \] - Compute \( [1 + x^{3/2}]^{-2} \): \[ [1 + 1]^{-2} = [2]^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4} \] Now, substitute these values into \( g'(x) \): \[ g'(1) = -\left( \dfrac{3}{2} \times 1 \right) \times \dfrac{1}{4} = -\dfrac{3}{2} \times \dfrac{1}{4} \] Simplify the expression: \[ g'(1) = -\dfrac{3}{2} \times \dfrac{1}{4} = -\dfrac{3}{8} \] **Answer:** \[ g'(1) = -\dfrac{3}{8} \] |
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| deepseekv3 | To find \( g'(x) \) when \( x = 1 \), we'll follow these steps: Given: \[ g(x) = \frac{1}{1 + x^{3/2}} \] ### Step 1: Rewrite \( g(x) \) using exponent notation \[ g(x) = \left(1 + x^{3/2}\right)^{-1} \] ### Step 2: Differentiate \( g(x) \) using the chain rule The chain rule states that if \( g(x) = f(u(x)) \), then \( g'(x) = f'(u(x)) \cdot u'(x) \). Here, \( f(u) = u^{-1} \) and \( u(x) = 1 + x^{3/2} \). First, compute \( f'(u) \): \[ f'(u) = -u^{-2} \] Next, compute \( u'(x) \): \[ u'(x) = \frac{d}{dx}\left(1 + x^{3/2}\right) = \frac{3}{2}x^{1/2} \] Now, apply the chain rule: \[ g'(x) = f'(u(x)) \cdot u'(x) = -\left(1 + x^{3/2}\right)^{-2} \cdot \frac{3}{2}x^{1/2} \] ### Step 3: Simplify \( g'(x) \) \[ g'(x) = -\frac{3}{2} \cdot \frac{x^{1/2}}{\left(1 + x^{3/2}\right)^2} \] ### Step 4: Evaluate \( g'(x) \) at \( x = 1 \) Substitute \( x = 1 \) into the expression: \[ g'(1) = -\frac{3}{2} \cdot \frac{1^{1/2}}{\left(1 + 1^{3/2}\right)^2} = -\frac{3}{2} \cdot \frac{1}{(1 + 1)^2} \] \[ g'(1) = -\frac{3}{2} \cdot \frac{1}{4} = -\frac{3}{8} \] ### Final Answer \[ g'(1) = -\frac{3}{8} \] |
-0.375 |