| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the second derivative of \(h(x) = \frac{1}{\sqrt{x} + 1}\) at \(x = 1\), we first need to find the first derivative \(h'(x)\) and then the second derivative \(h''(x)\). Given \(h(x) = \frac{1}{\sqrt{x} + 1}\), let's start by finding \(h'(x)\). First, rewrite \(h(x)\) as: \[h(x) = (x^{1/2} + 1)^{-1}\] Using the chain rule, we find the derivative of \(h(x)\) with respect to \(x\): \[ h'(x) = -1(x^{1/2} + 1)^{-2} \cdot \frac{1}{2}x^{-1/2} \] Simplifying: \[ h'(x) = -\frac{1}{2(x^{1/2} + 1)^2} \cdot \frac{1}{\sqrt{x}} \] \[ h'(x) = -\frac{1}{2\sqrt{x}(x^{1/2} + 1)^2} \] Next, we find the second derivative \(h''(x)\) by differentiating \(h'(x)\) with respect to \(x\): \[ h''(x) = \frac{d}{dx}\left(-\frac{1}{2\sqrt{x}(x^{1/2} + 1)^2}\right) \] Using the quotient rule and chain rule: \[ h''(x) = -\frac{d}{dx}\left(\frac{1}{2\sqrt{x}(x^{1/2} + 1)^2}\right) \] \[ = -\frac{2\sqrt{x} \cdot (-\frac{1}{2}x^{-3/2}(x^{1/2} + 1)^2) - 1 \cdot 2(x^{1/2} + 1)\cdot \frac{1}{2}x^{-1/2}}{4x(x^{1/2} + 1)^4} \] \[ = \frac{x^{-1/2}(x^{1/2} + 1)^2 + (x^{1/2} + 1)x^{-1/2}}{4x(x^{1/2} + 1)^4} \] \[ = \frac{x^{-1/2}(x^{1/2} + 1)(x^{1/2} + 1 + x^{1/2})}{4x(x^{1/2} + 1)^4} \] \[ = \frac{2x^{-1/2}(x^{1/2} + 1)}{4x(x^{1/2} + 1)^4} \] \[ = \frac{1}{2x^{3/2}(x^{1/2} + 1)^3} \] Now, we substitute \(x = 1\) into \(h''(x)\) to find \(h''(1)\): \[ h''(1) = \frac{1}{2 \cdot 1^{3/2} \cdot (1^{1/2} + 1)^3} \] \[ = \frac{1}{2 \cdot 1 \cdot (1 + 1)^3} \] \[ = \frac{1}{2 \cdot 2^3} \] \[ = \frac{1}{16} \] Therefore, \(h''(1) = \frac{1}{16}\). |
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| deepseekr1 | To find the second derivative of \( h(x) = \frac{1}{\sqrt{x} + 1} \) at \( x = 1 \), we start by computing the first derivative using the quotient rule or chain rule. First, we rewrite \( h(x) \) as \( (\sqrt{x} + 1)^{-1} \). Applying the chain rule, we get: \[ h'(x) = -(\sqrt{x} + 1)^{-2} \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x} (\sqrt{x} + 1)^2} \] Next, we find the second derivative \( h''(x) \). We use the quotient rule on \( h'(x) = -\frac{1}{2\sqrt{x} (\sqrt{x} + 1)^2} \). Let \( g(x) = 2\sqrt{x} (\sqrt{x} + 1)^2 \). The derivative \( g'(x) \) is computed using the product rule: \[ g'(x) = \frac{d}{dx} \left( 2\sqrt{x} (\sqrt{x} + 1)^2 \right) \] \[ = 2 \left( \frac{1}{2\sqrt{x}} (\sqrt{x} + 1)^2 + \sqrt{x} \cdot 2(\sqrt{x} + 1) \cdot \frac{1}{2\sqrt{x}} \right) \] \[ = \frac{(\sqrt{x} + 1)^2}{\sqrt{x}} + 2(\sqrt{x} + 1) \] \[ = (\sqrt{x} + 1) \left( \frac{\sqrt{x} + 1}{\sqrt{x}} + 2 \right) \] \[ = (\sqrt{x} + 1) \left( 1 + \frac{1}{\sqrt{x}} + 2 \right) \] \[ = (\sqrt{x} + 1) \left( 3 + \frac{1}{\sqrt{x}} \right) \] \[ = 3\sqrt{x} + 4 + \frac{1}{\sqrt{x}} \] Evaluating \( g'(x) \) at \( x = 1 \): \[ g'(1) = 3\sqrt{1} + 4 + \frac{1}{\sqrt{1}} = 3 + 4 + 1 = 8 \] The denominator \( g(x) \) at \( x = 1 \) is: \[ g(1) = 2\sqrt{1} (\sqrt{1} + 1)^2 = 2 \cdot 1 \cdot 4 = 8 \] Thus, the second derivative \( h''(x) \) at \( x = 1 \) is: \[ h''(1) = \frac{g'(1)}{[g(1)]^2} = \frac{8}{8^2} = \frac{8}{64} = \frac{1}{8} \] The value of \( h''(1) \) is \boxed{\dfrac{1}{8}}. |
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| gpto1 | To find \( h''(1) \) for \( h(x) = \dfrac{1}{\sqrt{x} + 1} \), we'll compute the first and second derivatives of \( h(x) \) and then evaluate at \( x = 1 \). **Step 1: Rewrite \( h(x) \) in Exponential Form** First, express \( h(x) \) using exponents to simplify differentiation: \[ h(x) = \dfrac{1}{x^{1/2} + 1} = [x^{1/2} + 1]^{-1} \] **Step 2: Compute the First Derivative \( h'(x) \)** Let \( f(x) = x^{1/2} + 1 \), so \( h(x) = [f(x)]^{-1} \). Using the chain rule: \[ h'(x) = -[f(x)]^{-2} \cdot f'(x) \] Compute \( f'(x) \): \[ f'(x) = \dfrac{1}{2} x^{-1/2} \] So, \[ h'(x) = -[x^{1/2} + 1]^{-2} \cdot \dfrac{1}{2} x^{-1/2} = -\dfrac{1}{2} x^{-1/2} [x^{1/2} + 1]^{-2} \] **Step 3: Compute the Second Derivative \( h''(x) \)** We'll differentiate \( h'(x) \): \[ h''(x) = -\dfrac{1}{2} \cdot \dfrac{d}{dx} \left( x^{-1/2} [x^{1/2} + 1]^{-2} \right) \] Let \( u(x) = x^{-1/2} \) and \( v(x) = [x^{1/2} + 1]^{-2} \). Then \( h''(x) \) becomes: \[ h''(x) = -\dfrac{1}{2} \left( u'(x)v(x) + u(x)v'(x) \right) \] Compute \( u'(x) \) and \( v'(x) \): \[ u'(x) = -\dfrac{1}{2} x^{-3/2}, \quad v'(x) = -2 [x^{1/2} + 1]^{-3} \cdot \dfrac{1}{2} x^{-1/2} = -[x^{1/2} + 1]^{-3} x^{-1/2} \] Now, compute \( h''(x) \): \[ h''(x) = -\dfrac{1}{2} \left( \left( -\dfrac{1}{2} x^{-3/2} \right) [x^{1/2} + 1]^{-2} + x^{-1/2} \left( -[x^{1/2} + 1]^{-3} x^{-1/2} \right) \right) \] Simplify the expression: \[ h''(x) = -\dfrac{1}{2} \left( -\dfrac{1}{2} x^{-3/2} [x^{1/2} + 1]^{-2} - x^{-1} [x^{1/2} + 1]^{-3} \right) \] Simplify signs: \[ h''(x) = \dfrac{1}{2} \left( \dfrac{1}{2} x^{-3/2} [x^{1/2} + 1]^{-2} + x^{-1} [x^{1/2} + 1]^{-3} \right) \] **Step 4: Evaluate \( h''(1) \)** Substitute \( x = 1 \): \[ x^{-3/2} = 1^{-3/2} = 1, \quad x^{-1} = 1^{-1} = 1, \quad x^{1/2} = 1^{1/2} = 1 \] Compute \( [x^{1/2} + 1] \): \[ [x^{1/2} + 1] = 1 + 1 = 2 \] Now compute each term: \[ h''(1) = \dfrac{1}{2} \left( \dfrac{1}{2} \cdot 1 \cdot (2)^{-2} + 1 \cdot (2)^{-3} \right) = \dfrac{1}{2} \left( \dfrac{1}{2} \cdot \dfrac{1}{4} + \dfrac{1}{8} \right) \] Simplify: \[ h''(1) = \dfrac{1}{2} \left( \dfrac{1}{8} + \dfrac{1}{8} \right) = \dfrac{1}{2} \cdot \dfrac{1}{4} = \dfrac{1}{8} \] **Answer:** \(\displaystyle h''(1)\;=\;\frac{1}{8}\) |
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| GPT-4o | To solve for \( h''(x) \) when \( x = 1 \), we need to find the second derivative of the function \( h(x) = \frac{1}{\sqrt{x} + 1} \) and then evaluate it at \( x = 1 \). ### Step-by-Step Solution: #### Step 1: Differentiate \( h(x) \) to find \( h'(x) \) First, let's rewrite \( h(x) \) in a form that is easier to differentiate: \[ h(x) = (\sqrt{x} + 1)^{-1} \] We can use the chain rule to find the first derivative \( h'(x) \). \[ h(x) = (u(x))^{-1} \] where \( u(x) = \sqrt{x} + 1 \). The derivative of \( (u(x))^{-1} \) with respect to \( x \) is: \[ h'(x) = -u(x)^{-2} \cdot u'(x) \] Now, we need to find \( u'(x) \): \[ u(x) = \sqrt{x} + 1 \] \[ u'(x) = \frac{d}{dx} (\sqrt{x}) + \frac{d}{dx} (1) \] \[ u'(x) = \frac{1}{2} x^{-1/2} + 0 \] \[ u'(x) = \frac{1}{2\sqrt{x}} \] Plugging \( u(x) \) and \( u'(x) \) back into the expression for \( h'(x) \): \[ h'(x) = -(\sqrt{x} + 1)^{-2} \cdot \frac{1}{2\sqrt{x}} \] Simplifying the expression: \[ h'(x) = -\frac{1}{2\sqrt{x}(\sqrt{x} + 1)^2} \] #### Step 2: Differentiate \( h'(x) \) to find \( h''(x) \) Next, we need to find the second derivative \( h''(x) \). Using the quotient rule, \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), let's first identify \( u \) and \( v \): \[ h'(x) = -\frac{1}{2\sqrt{x}(\sqrt{x} + 1)^2} \] Let \( u = 1 \) and \( v = 2\sqrt{x}(\sqrt{x} + 1)^2 \) Now, differentiate \( u \) and \( v \): \[ u' = 0 \] \[ v = 2\sqrt{x}(\sqrt{x} + 1)^2 \] Use the product and chain rule to find \( v' \): \[ v = 2\sqrt{x}(\sqrt{x} + 1)^2 \] \[ v' = 2 \left( \frac{d}{dx} (\sqrt{x}) \cdot (\sqrt{x} + 1)^2 + \sqrt{x} \cdot \frac{d}{dx}((\sqrt{x} + 1)^2) \right) \] \[ v' = 2 \left( \frac{1}{2\sqrt{x}} (\sqrt{x} + 1)^2 + \sqrt{x} \cdot 2(\sqrt{x} + 1) \cdot \frac{1}{2\sqrt{x}} \right) \] \[ v' = 2 \left( \frac{1}{2\sqrt{x}} (\sqrt{x} + 1)^2 + (\sqrt{x} + 1) \right) \] \[ v' = (\sqrt{x} + 1)^2 \cdot \frac{1}{\sqrt{x}} + 2(\sqrt{x} + 1) \] \[ v' = \frac{(\sqrt{x} + 1)^2}{\sqrt{x}} + 2(\sqrt{x} + 1) \] \[ v' = \frac{(\sqrt{x} + 1)^2 + 2\sqrt{x}(\sqrt{x} + 1)}{\sqrt{x}} \] \[ v' = \frac{(\sqrt{x} + 1)(\sqrt{x} + 1 + 2\sqrt{x})}{\sqrt{x}} \] \[ v' = \frac{(\sqrt{x} + 1)(3\sqrt{x} + 1)}{\sqrt{x}} \] Now apply the quotient rule: \[ h''(x) = \frac{u'v - uv'}{v^2} \] \[ h''(x) = \frac{0 \cd |
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| llama3.1-70B-Instruct | To find h''(x) when x = 1, we need to find the second derivative of h(x) and then evaluate it at x = 1. First, let's find the first derivative of h(x): h(x) = 1/(\sqrt{x} + 1) Using the quotient rule, we get: h'(x) = (-1/(\sqrt{x} + 1)^2) \* (1/2\sqrt{x}) Simplifying, we get: h'(x) = -1/(2\sqrt{x}(\sqrt{x} + 1)^2) Now, let's find the second derivative of h(x): h''(x) = d/dx (-1/(2\sqrt{x}(\sqrt{x} + 1)^2)) Using the product rule and chain rule, we get: h''(x) = (1/(2\sqrt{x}(\sqrt{x} + 1)^2)) \* (1/2\sqrt{x}) - (1/(2\sqrt{x}(\sqrt{x} + 1)^2)) \* (2/(\sqrt{x} + 1)) Simplifying, we get: h''(x) = 1/(4x\sqrt{x}(\sqrt{x} + 1)^2) - 2/(2\sqrt{x}(\sqrt{x} + 1)^3) Now, let's evaluate h''(x) at x = 1: h''(1) = 1/(4(1)\sqrt{1}(1 + 1)^2) - 2/(2\sqrt{1}(1 + 1)^3) Simplifying, we get: h''(1) = 1/16 - 1/8 h''(1) = -1/16 Therefore, h''(1) = -1/16. |
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| llama3.1-405B-Instruct | To find h''(x), we need to find the second derivative of h(x). First, let's find the first derivative of h(x): h'(x) = d(1/(\sqrt{x} + 1))/dx Using the quotient rule, we get: h'(x) = -1/(\sqrt{x} + 1)^2 \* (1/2\*x^(-1/2)) Simplifying, we get: h'(x) = -1/(2\*x^(3/2) + 2\*x + 2\*sqrt(x)) Now, let's find the second derivative of h(x): h''(x) = d(-1/(2\*x^(3/2) + 2\*x + 2\*sqrt(x)))/dx Using the quotient rule again, we get: h''(x) = 1/(2\*x^(3/2) + 2\*x + 2\*sqrt(x))^2 \* (3\*x^(1/2) + 2 + 1/sqrt(x)) Simplifying, we get: h''(x) = (3\*x^(1/2) + 2 + 1/sqrt(x))/(4\*x^3 + 8\*x^(5/2) + 8\*x^2 + 8\*x^(3/2) + 4\*x + 4\*sqrt(x)) Now, we need to find h''(1): h''(1) = (3\*1^(1/2) + 2 + 1/sqrt(1))/(4\*1^3 + 8\*1^(5/2) + 8\*1^2 + 8\*1^(3/2) + 4\*1 + 4\*sqrt(1)) Simplifying, we get: h''(1) = (3 + 2 + 1)/(4 + 8 + 8 + 8 + 4 + 4) h''(1) = 6/36 h''(1) = 1/6 Therefore, h''(1) = 1/6. |
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| DeepSeek-Coder-V2 | To find h''(x) when x = 1, we need to follow these steps: 1. Find the first derivative h'(x) of the function h(x) = 1/(\sqrt{x} + 1). 2. Find the second derivative h''(x) of the function h(x). 3. Evaluate h''(x) at x = 1. Step 1: Find the first derivative h'(x) h(x) = 1/(\sqrt{x} + 1) To find the first derivative, we can use the quotient rule, which states that if we have a function f(x) = u(x)/v(x), then the derivative f'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2. In this case, u(x) = 1 and v(x) = \sqrt{x} + 1. u'(x) = 0 (since the derivative of a constant is 0) v'(x) = d/dx (\sqrt{x} + 1) = 1/(2\sqrt{x}) Now, apply the quotient rule: h'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2 h'(x) = (0 * (\sqrt{x} + 1) - 1 * (1/(2\sqrt{x}))) / ((\sqrt{x} + 1))^2 h'(x) = (-1/(2\sqrt{x})) / ((\sqrt{x} + 1))^2 h'(x) = -1 / (2\sqrt{x} * (\sqrt{x} + 1)^2) Step 2: Find the second derivative h''(x) Now, we need to find the second derivative of h(x). We will use the quotient rule again, but this time, we will apply it to h'(x). Let u(x) = -1 and v(x) = 2\sqrt{x} * (\sqrt{x} + 1)^2 u'(x) = 0 v'(x) = d/dx (2\sqrt{x} * (\sqrt{x} + 1)^2) To find v'(x), we will use the product rule and the chain rule: v'(x) = 2 * (d/dx (\sqrt{x}) * (\sqrt{x} + 1)^2 + \sqrt{x} * d/dx ((\sqrt{x} + 1)^2)) v'(x) = 2 * (1/(2\sqrt{x}) * (\sqrt{x} + 1)^2 + \sqrt{x} * (2 * (\sqrt{x} + 1) * 1/(2\sqrt{x}))) v'(x) = 2 * (1/(2\sqrt{x}) * (\sqrt{x} + 1)^2 + (\sqrt{x} + 1)) v'(x) = (\sqrt{x} + 1)^2 / \sqrt{x} + 2(\sqrt{x} + 1) Now, apply the quotient rule to h'(x): h''(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2 h''(x) = (0 * (2\sqrt{x} * (\sqrt{x} + 1)^2) - (-1) * ((\sqrt{x} + 1)^2 / \sqrt{x} + 2(\sqrt{x} + 1))) / (2\sqrt{x} * (\sqrt{x} + 1)^2)^2 h''(x) = ((\sqrt{x} + 1)^2 / \sqrt{x} + 2(\sqrt{x} + 1)) / (4x * (\sqrt{x} + 1)^4) Step 3: Evaluate h''(x) at x = 1 Now, we need to evaluate h''(x) at x = 1: h''(1) = ((\sqrt{1} + 1)^2 / \sqrt{1} + 2(\sqrt{1} + 1)) / (4 * 1 * (\sqrt{1} + 1)^4) h''(1) = ((1 + 1)^2 / 1 + 2(1 + 1)) / (4 * (1 + 1)^4) h''(1) = (4 / 1 + 4) / (4 * 16) h''(1) = (4 + 4) / 64 h''(1) = 8 / 64 h''(1) = 1 / 8 So, h''(x) when x = 1 is 1/8. |
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| Qwen2-72B-Instruct | To find the second derivative of \(h(x) = \frac{1}{\sqrt{x} + 1}\) at \(x = 1\), we first need to find the first derivative \(h'(x)\) and then the second derivative \(h''(x)\). Given \(h(x) = \frac{1}{\sqrt{x} + 1}\), let's start by finding \(h'(x)\). First, rewrite \(h(x)\) as: \[h(x) = (x^{1/2} + 1)^{-1}\] Using the chain rule, we find the derivative of \(h(x)\) with respect to \(x\): \[ h'(x) = -1(x^{1/2} + 1)^{-2} \cdot \frac{1}{2}x^{-1/2} \] Simplifying: \[ h'(x) = -\frac{1}{2(x^{1/2} + 1)^2} \cdot \frac{1}{\sqrt{x}} \] \[ h'(x) = -\frac{1}{2\sqrt{x}(x^{1/2} + 1)^2} \] Next, we find the second derivative \(h''(x)\) by differentiating \(h'(x)\) with respect to \(x\): \[ h''(x) = \frac{d}{dx}\left(-\frac{1}{2\sqrt{x}(x^{1/2} + 1)^2}\right) \] Using the quotient rule and chain rule: \[ h''(x) = -\frac{d}{dx}\left(\frac{1}{2\sqrt{x}(x^{1/2} + 1)^2}\right) \] \[ = -\frac{2\sqrt{x} \cdot (-\frac{1}{2}x^{-3/2}(x^{1/2} + 1)^2) - 1 \cdot 2(x^{1/2} + 1)\cdot \frac{1}{2}x^{-1/2}}{4x(x^{1/2} + 1)^4} \] \[ = \frac{x^{-1/2}(x^{1/2} + 1)^2 + (x^{1/2} + 1)x^{-1/2}}{4x(x^{1/2} + 1)^4} \] \[ = \frac{x^{-1/2}(x^{1/2} + 1)(x^{1/2} + 1 + x^{1/2})}{4x(x^{1/2} + 1)^4} \] \[ = \frac{2x^{-1/2}(x^{1/2} + 1)}{4x(x^{1/2} + 1)^4} \] \[ = \frac{1}{2x^{3/2}(x^{1/2} + 1)^3} \] Now, we substitute \(x = 1\) into \(h''(x)\) to find \(h''(1)\): \[ h''(1) = \frac{1}{2 \cdot 1^{3/2} \cdot (1^{1/2} + 1)^3} \] \[ = \frac{1}{2 \cdot 1 \cdot (1 + 1)^3} \] \[ = \frac{1}{2 \cdot 2^3} \] \[ = \frac{1}{16} \] Therefore, \(h''(1) = \frac{1}{16}\). |
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| baichuan4 | To find the second derivative of \(h(x) = \frac{1}{\sqrt{x} + 1}\), we first need to find the first derivative \(h'(x)\). Using the quotient rule for derivatives, we have: \[h'(x) = \frac{-1 \cdot \frac{1}{2}x^{-\frac{1}{2}} \cdot (\sqrt{x} + 1) - (-1) \cdot 1}{(\sqrt{x} + 1)^2}\] \[= \frac{-\frac{1}{2}x^{-\frac{1}{2}}(\sqrt{x} + 1) + 1}{(\sqrt{x} + 1)^2}\] \[= \frac{-\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x} - \frac{1}{2}x^{-\frac{1}{2}} + 1}{(\sqrt{x} + 1)^2}\] \[= \frac{-\frac{1}{2} - \frac{1}{2}x^{-\frac{1}{2}} + 1}{(\sqrt{x} + 1)^2}\] \[= \frac{1 - \frac{1}{2} - \frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x} + 1)^2}\] \[= \frac{\frac{1}{2} - \frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x} + 1)^2}\] \[= \frac{1 - x^{-\frac{1}{2}}}{2(\sqrt{x} + 1)^2}\] Now, we need to find the second derivative \(h''(x)\). We will use the quotient rule again: \[h''(x) = \frac{(1 \cdot 2(\sqrt{x} + 1)^2) \cdot (-\frac{1}{2}x^{-\frac{3}{2}}) - (1 - x^{-\frac{1}{2}}) \cdot (2(\sqrt{x} + 1) \cdot \frac{1}{2}x^{-\frac{1}{2}})}{(2(\sqrt{x} + 1)^2)^2}\] \[= \frac{-2(\sqrt{x} + 1)^2 \cdot \frac{1}{2}x^{-\frac{3}{2}} - (1 - x^{-\frac{1}{2}}) \cdot (\sqrt{x} + 1) \cdot x^{-\frac{1}{2}}}{4(\sqrt{x} + 1)^4}\] \[= \frac{-(\sqrt{x} + 1)^2 \cdot x^{-\frac{3}{2}} - (1 - x^{-\frac{1}{2}}) \cdot (\sqrt{x} + 1) \cdot x^{-\frac{1}{2}}}{4(\sqrt{x} + 1)^4}\] Now, let's plug in \(x = 1\): \[h''(1) = \frac{-(\sqrt{1} + 1)^2 \cdot 1^{-\frac{3}{2}} - (1 - 1^{-\frac{1}{2}}) \cdot (\sqrt{1} + 1) \cdot 1^{-\frac{1}{2}}}{4(\sqrt{1} + 1)^4}\] \[= \frac{-(1 + 1)^2 \cdot 1 - (1 - 1) \cdot (1 + 1) \cdot 1}{4(1 + 1)^4}\] \[= \frac{-4 \cdot 1 - 0 \cdot 2 \cdot 1}{4 \cdot 16}\] \[= \frac{-4}{64}\] \[= -\frac{1}{16}\] So, \(h''(1) = -\frac{1}{16}\). |
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| doubao | First find the first derivative h'(x): \[ \begin{align*} h'(x)&=(\frac{1}{\sqrt{x}+1})'\\ &=-\frac{1}{2}(\sqrt{x}+1)^{-2}\times\frac{1}{2\sqrt{x}}\\ &=-\frac{1}{4\sqrt{x}(\sqrt{x}+1)^{2}} \end{align*} \] Then find the second derivative h''(x): \[ \begin{align*} h''(x)&=(-\frac{1}{4\sqrt{x}(\sqrt{x}+1)^{2}})'\\ &=-\frac{1}{4}[(\sqrt{x})'(\sqrt{x}+1)^{2}+\sqrt{x}((\sqrt{x}+1)^{2})']\\ &=-\frac{1}{4}[\frac{1}{2\sqrt{x}}(\sqrt{x}+1)^{2}+\sqrt{x}\times2(\sqrt{x}+1)\times\frac{1}{2\sqrt{x}}]\\ &=-\frac{1}{4}[\frac{(\sqrt{x}+1)^{2}}{2\sqrt{x}}+\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}}]\\ &=-\frac{1}{4}[\frac{x+2\sqrt{x}+1}{2\sqrt{x}}+\sqrt{x}+1]\\ &=-\frac{1}{4}[\frac{x+2\sqrt{x}+1+2\sqrt{x}( \sqrt{x}+1)}{2\sqrt{x}}]\\ &=-\frac{1}{4}[\frac{x+2\sqrt{x}+1+2x+2\sqrt{x}}{2\sqrt{x}}]\\ &=-\frac{1}{4}[\frac{3x+4\sqrt{x}+1}{2\sqrt{x}}] \end{align*} \] When x=1, \[ \begin{align*} h''(1)&=-\frac{1}{4}[\frac{3\times1+4\sqrt{1}+1}{2\sqrt{1}}]\\ &=-\frac{1}{4}[\frac{3+4+1}{2}]\\ &=-\frac{1}{4}\times4\\ &=-1 \end{align*} \] |
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| minimax | To find the second derivative of the function \( h(x) = \frac{1}{\sqrt{x} + 1} \) at \( x = 1 \), we first need to find the first derivative \( h'(x) \) and then the second derivative \( h''(x) \). Let's start by finding the first derivative \( h'(x) \). We can use the chain rule and the quotient rule for differentiation. The chain rule states that if you have a composite function \( f(g(x)) \), then its derivative is \( f'(g(x)) \cdot g'(x) \). The quotient rule states that if you have a function \( \frac{u(x)}{v(x)} \), then its derivative is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] Let's apply the quotient rule to \( h(x) \), where \( u(x) = 1 \) and \( v(x) = \sqrt{x} + 1 \). The derivatives of \( u \) and \( v \) are \( u'(x) = 0 \) and \( v'(x) = \frac{1}{2\sqrt{x}} \), respectively. Applying the quotient rule: \[ h'(x) = \frac{0 \cdot (\sqrt{x} + 1) - 1 \cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x} + 1)^2} \] \[ h'(x) = \frac{-\frac{1}{2\sqrt{x}}}{(\sqrt{x} + 1)^2} \] \[ h'(x) = \frac{-1}{2x\sqrt{x} + 2\sqrt{x}} \] Now, let's find the second derivative \( h''(x) \). We'll differentiate \( h'(x) \) using the quotient rule again. Let \( u(x) = -1 \) and \( v(x) = 2x\sqrt{x} + 2\sqrt{x} \). The derivatives of \( u \) and \( v \) are \( u'(x) = 0 \) and \( v'(x) = 2\sqrt{x} + \frac{1}{\sqrt{x}} \), respectively. Applying the quotient rule: \[ h''(x) = \frac{0 \cdot (2x\sqrt{x} + 2\sqrt{x}) - (-1) \cdot (2\sqrt{x} + \frac{1}{\sqrt{x}})}{(2x\sqrt{x} + 2\sqrt{x})^2} \] \[ h''(x) = \frac{2\sqrt{x} + \frac{1}{\sqrt{x}}}{(2x\sqrt{x} + 2\sqrt{x})^2} \] \[ h''(x) = \frac{2\sqrt{x} + \frac{1}{\sqrt{x}}}{4x^2x + 8x\sqrt{x} + 4x} \] \[ h''(x) = \frac{2\sqrt{x} + \frac{1}{\sqrt{x}}}{4x^{2.5} + 8x^{1.5} + 4x} \] Now we need to evaluate \( h''(x) \) at \( x = 1 \): \[ h''(1) = \frac{2\sqrt{1} + \frac{1}{\sqrt{1}}}{4(1)^{2.5} + 8(1)^{1.5} + 4(1)} \] \[ h''(1) = \frac{2 + 1}{4 + 8 + 4} \] \[ h''(1) = \frac{3}{16} \] So, the second derivative of \( h(x) \) when \( x = 1 \) is \( \frac{3}{16} \). |
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| kimi | To find the second derivative \( h''(x) \) of the function \( h(x) = \frac{1}{\sqrt{x} + 1} \) at \( x = 1 \), we first need to find the first derivative \( h'(x) \) and then differentiate it again to get \( h''(x) \). Let's start by finding \( h'(x) \): Given: \[ h(x) = \frac{1}{\sqrt{x} + 1} \] We can rewrite \( h(x) \) as: \[ h(x) = (\sqrt{x} + 1)^{-1} \] Now, apply the chain rule to find the first derivative: \[ h'(x) = -1 \cdot (\sqrt{x} + 1)^{-2} \cdot \frac{1}{2\sqrt{x}} \] \[ h'(x) = -\frac{1}{2(\sqrt{x} + 1)^2 \sqrt{x}} \] Now, let's find \( h''(x) \): \[ h''(x) = \frac{d}{dx} \left( -\frac{1}{2(\sqrt{x} + 1)^2 \sqrt{x}} \right) \] We'll use the product rule and chain rule here. Let \( u = -\frac{1}{2} \) and \( v = (\sqrt{x} + 1)^2 \sqrt{x} \), then \( h'(x) = u \cdot v \). \[ h''(x) = u' \cdot v + u \cdot v' \] First, find \( u' \) and \( v' \): \[ u' = 0 \] (since \( u \) is a constant) For \( v' \), we'll use the chain rule again: \[ v = (\sqrt{x} + 1)^2 \sqrt{x} \] Let \( w = \sqrt{x} \), then \( v = (w + 1)^2 w \) \[ v' = 2(w + 1)w \cdot \frac{1}{2\sqrt{x}} + (w + 1)^2 \cdot \frac{1}{2\sqrt{x}} \] \[ v' = (w + 1)w \cdot \frac{1}{\sqrt{x}} + (w + 1)^2 \cdot \frac{1}{2\sqrt{x}} \] Substitute \( w = \sqrt{x} \) back into \( v' \): \[ v' = (\sqrt{x} + 1)\sqrt{x} \cdot \frac{1}{\sqrt{x}} + (\sqrt{x} + 1)^2 \cdot \frac{1}{2\sqrt{x}} \] \[ v' = (\sqrt{x} + 1) + \frac{(\sqrt{x} + 1)^2}{2\sqrt{x}} \] Now, substitute \( u' \) and \( v' \) back into the expression for \( h''(x) \): \[ h''(x) = 0 \cdot (\sqrt{x} + 1)^2 \sqrt{x} + \left( -\frac{1}{2(\sqrt{x} + 1)^2 \sqrt{x}} \right) \left( (\sqrt{x} + 1) + \frac{(\sqrt{x} + 1)^2}{2\sqrt{x}} \right) \] Simplify the expression: \[ h''(x) = -\frac{1}{2(\sqrt{x} + 1)^2 \sqrt{x}} \left( \frac{(\sqrt{x} + 1)^2 \sqrt{x} + (\sqrt{x} + 1)^3}{2\sqrt{x}} \right) \] \[ h''(x) = -\frac{(\sqrt{x} + 1)^2 + (\sqrt{x} + 1)^3}{4(\sqrt{x} + 1)^2 x} \] Now, we can find \( h''(1) \): \[ h''(1) = -\frac{(\sqrt{1} + 1)^2 + (\sqrt{1} + 1)^3}{4(\sqrt{1} + 1)^2 \cdot 1} \] \[ h''(1) = -\frac{(1 + 1)^2 + (1 + 1)^3}{4(1 + 1)^2} \] \[ h''(1) = |
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| wenxinyiyan | 为了找到h''(x)当x=1时的值，我们首先需要找到h(x)的一阶和二阶导数。 首先，给定函数 $h(x) = \frac{1}{\sqrt{x} + 1}$。 步骤1：求h'(x) 使用求导的链式法则和商的求导法则，我们有： $h'(x) = -\frac{1}{(\sqrt{x}+1)^2} \cdot \frac{d}{dx}(\sqrt{x}+1)$ 由于 $\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}$，我们得到： $h'(x) = -\frac{1}{(\sqrt{x}+1)^2} \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x}(\sqrt{x}+1)^2}$ 步骤2：求h''(x) 继续使用求导的链式法则和商的求导法则，我们对h'(x)求导得到h''(x)： $h''(x) = \frac{d}{dx}\left(-\frac{1}{2\sqrt{x}(\sqrt{x}+1)^2}\right)$ 这个导数计算比较复杂，但我们可以分步来做。首先，令 $u(x) = 2\sqrt{x}(\sqrt{x}+1)^2$，则 $h'(x) = -\frac{1}{u(x)}$。 现在我们需要找到u'(x)： $u'(x) = \frac{d}{dx}(2\sqrt{x}(\sqrt{x}+1)^2)$ 使用乘法法则和链式法则，我们得到： $u'(x) = 2\sqrt{x} \cdot 2(\sqrt{x}+1) \cdot \frac{1}{2\sqrt{x}} + (\sqrt{x}+1)^2 \cdot \frac{1}{2\sqrt{x}}$ $= 4(\sqrt{x}+1) + \frac{(\sqrt{x}+1)^2}{2\sqrt{x}}$ 现在我们可以找到h''(x)： $h''(x) = -\frac{-u'(x)}{u(x)^2}$ $= \frac{4(\sqrt{x}+1) + \frac{(\sqrt{x}+1)^2}{2\sqrt{x}}}{(2\sqrt{x}(\sqrt{x}+1)^2)^2}$ 步骤3：求h''(1) 将x=1代入h''(x)的表达式中，我们得到： $h''(1) = \frac{4(\sqrt{1}+1) + \frac{(\sqrt{1}+1)^2}{2\sqrt{1}}}{(2\sqrt{1}(\sqrt{1}+1)^2)^2}$ $= \frac{4(1+1) + \frac{(1+1)^2}{2}}{(2(1+1)^2)^2}$ $= \frac{8 + 2}{64}$ $= \frac{5}{32}$ |
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| zhipu | To find the second derivative \( h''(x) \) of the function \( h(x) = \frac{1}{\sqrt{x} + 1} \), we need to differentiate \( h(x) \) twice. Let's start by finding the first derivative \( h'(x) \). ### Step 1: Find \( h'(x) \) We can use the quotient rule for the first derivative, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then: \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \] In our case, \( g(x) = 1 \) and \( h(x) = \sqrt{x} + 1 \). So, we have: \[ g'(x) = 0 \] \[ h'(x) = \frac{d}{dx}(\sqrt{x} + 1) = \frac{1}{2\sqrt{x}} \] Applying the quotient rule: \[ h'(x) = \frac{0 \cdot (\sqrt{x} + 1) - 1 \cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x} + 1)^2} \] \[ h'(x) = -\frac{1}{2\sqrt{x}(\sqrt{x} + 1)^2} \] ### Step 2: Find \( h''(x) \) Now, we need to differentiate \( h'(x) \) to find \( h''(x) \). We can use the product rule and the chain rule for this. Recall that the product rule states: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] Let \( u(x) = -\frac{1}{2\sqrt{x}} \) and \( v(x) = (\sqrt{x} + 1)^{-2} \). First, find \( u'(x) \) and \( v'(x) \): \[ u'(x) = \frac{d}{dx}\left(-\frac{1}{2\sqrt{x}}\right) = \frac{1}{4x\sqrt{x}} \] \[ v'(x) = \frac{d}{dx}\left((\sqrt{x} + 1)^{-2}\right) = -2(\sqrt{x} + 1)^{-3} \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{\sqrt{x}(\sqrt{x} + 1)^3} \] Now, apply the product rule: \[ h''(x) = u'(x)v(x) + u(x)v'(x) \] \[ h''(x) = \left(\frac{1}{4x\sqrt{x}}\right)\left((\sqrt{x} + 1)^{-2}\right) + \left(-\frac{1}{2\sqrt{x}}\right)\left(-\frac{1}{\sqrt{x}(\sqrt{x} + 1)^3}\right) \] \[ h''(x) = \frac{1}{4x\sqrt{x}(\sqrt{x} + 1)^2} + \frac{1}{2x\sqrt{x}(\sqrt{x} + 1)^3} \] ### Step 3: Simplify and Evaluate at \( x = 1 \) Combine the terms under a common denominator: \[ h''(x) = \frac{1}{4x\sqrt{x}(\sqrt{x} + 1)^2} + \frac{1}{2x\sqrt{x}(\sqrt{x} + 1)^3} \] \[ h''(x) = \frac{(\sqrt{x} + 1) + 2}{4x\sqrt{x}(\sqrt{x} + 1)^3} \] \[ h''(x) = \frac{\sqrt{x} + 3}{4x\sqrt{x}(\sqrt{x} + 1)^3} \] Now, evaluate \( h''(x) \) at \( x = 1 \): \[ h''(1) = \frac{\sqrt{1} + 3}{4 \cdot 1 \cdot \sqrt{1} \cdot (\sqrt{1} + 1)^3} \] \[ h''(1) = \frac{1 + 3}{4 \cdot 1 \cdot 1 \cdot 2^3} \] \[ h''(1) = \frac{4}{4 \cdot 8} \] \[ h''(1) = \frac{4}{32} \] \[ h''(1) = \frac{1}{8} \] Therefore, \( h''(1) = \frac{1}{8} \). |
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| gpto1 | To find \( h''(1) \) for \( h(x) = \dfrac{1}{\sqrt{x} + 1} \), we'll compute the first and second derivatives of \( h(x) \) and then evaluate at \( x = 1 \). **Step 1: Rewrite \( h(x) \) in Exponential Form** First, express \( h(x) \) using exponents to simplify differentiation: \[ h(x) = \dfrac{1}{x^{1/2} + 1} = [x^{1/2} + 1]^{-1} \] **Step 2: Compute the First Derivative \( h'(x) \)** Let \( f(x) = x^{1/2} + 1 \), so \( h(x) = [f(x)]^{-1} \). Using the chain rule: \[ h'(x) = -[f(x)]^{-2} \cdot f'(x) \] Compute \( f'(x) \): \[ f'(x) = \dfrac{1}{2} x^{-1/2} \] So, \[ h'(x) = -[x^{1/2} + 1]^{-2} \cdot \dfrac{1}{2} x^{-1/2} = -\dfrac{1}{2} x^{-1/2} [x^{1/2} + 1]^{-2} \] **Step 3: Compute the Second Derivative \( h''(x) \)** We'll differentiate \( h'(x) \): \[ h''(x) = -\dfrac{1}{2} \cdot \dfrac{d}{dx} \left( x^{-1/2} [x^{1/2} + 1]^{-2} \right) \] Let \( u(x) = x^{-1/2} \) and \( v(x) = [x^{1/2} + 1]^{-2} \). Then \( h''(x) \) becomes: \[ h''(x) = -\dfrac{1}{2} \left( u'(x)v(x) + u(x)v'(x) \right) \] Compute \( u'(x) \) and \( v'(x) \): \[ u'(x) = -\dfrac{1}{2} x^{-3/2}, \quad v'(x) = -2 [x^{1/2} + 1]^{-3} \cdot \dfrac{1}{2} x^{-1/2} = -[x^{1/2} + 1]^{-3} x^{-1/2} \] Now, compute \( h''(x) \): \[ h''(x) = -\dfrac{1}{2} \left( \left( -\dfrac{1}{2} x^{-3/2} \right) [x^{1/2} + 1]^{-2} + x^{-1/2} \left( -[x^{1/2} + 1]^{-3} x^{-1/2} \right) \right) \] Simplify the expression: \[ h''(x) = -\dfrac{1}{2} \left( -\dfrac{1}{2} x^{-3/2} [x^{1/2} + 1]^{-2} - x^{-1} [x^{1/2} + 1]^{-3} \right) \] Simplify signs: \[ h''(x) = \dfrac{1}{2} \left( \dfrac{1}{2} x^{-3/2} [x^{1/2} + 1]^{-2} + x^{-1} [x^{1/2} + 1]^{-3} \right) \] **Step 4: Evaluate \( h''(1) \)** Substitute \( x = 1 \): \[ x^{-3/2} = 1^{-3/2} = 1, \quad x^{-1} = 1^{-1} = 1, \quad x^{1/2} = 1^{1/2} = 1 \] Compute \( [x^{1/2} + 1] \): \[ [x^{1/2} + 1] = 1 + 1 = 2 \] Now compute each term: \[ h''(1) = \dfrac{1}{2} \left( \dfrac{1}{2} \cdot 1 \cdot (2)^{-2} + 1 \cdot (2)^{-3} \right) = \dfrac{1}{2} \left( \dfrac{1}{2} \cdot \dfrac{1}{4} + \dfrac{1}{8} \right) \] Simplify: \[ h''(1) = \dfrac{1}{2} \left( \dfrac{1}{8} + \dfrac{1}{8} \right) = \dfrac{1}{2} \cdot \dfrac{1}{4} = \dfrac{1}{8} \] **Answer:** \(\displaystyle h''(1)\;=\;\frac{1}{8}\) |
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| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-d6e27f97-e353-948f-9eb5-a54934a704bf', 'request_id': 'd6e27f97-e353-948f-9eb5-a54934a704bf'} | 0.125 |