| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the derivative \(f'(x)\) of the given implicit function \(x^2y^2 - 3y + 2x^3 = 0\) with respect to \(x\), we need to perform implicit differentiation. After finding the derivative, we can then evaluate \(f'(1)\) at the points \((1,1)\) and \((1,2)\). Let's differentiate the equation with respect to \(x\): \[ \frac{d}{dx}(x^2y^2) - \frac{d}{dx}(3y) + \frac{d}{dx}(2x^3) = \frac{d}{dx}(0) \] Applying the product rule and chain rule where necessary: \[ 2xy^2 + x^2(2y\frac{dy}{dx}) - 3\frac{dy}{dx} + 6x^2 = 0 \] Rearranging for \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(2x^2y - 3) = -2xy^2 - 6x^2 \] Thus, \[ \frac{dy}{dx} = \frac{-2xy^2 - 6x^2}{2x^2y - 3} \] When \(x = 1\), the expression for \(\frac{dy}{dx}\) becomes: \[ \frac{dy}{dx} = \frac{-2y^2 - 6}{2y - 3} \] Now we can evaluate \(\frac{dy}{dx}\) at the points \((1,1)\) and \((1,2)\): 1. At \((1,1)\): \[ \frac{dy}{dx} = \frac{-2(1)^2 - 6}{2(1) - 3} = \frac{-8}{-1} = 8 \] 2. At \((1,2)\): \[ \frac{dy}{dx} = \frac{-2(2)^2 - 6}{2(2) - 3} = \frac{-10}{1} = -10 \] Therefore, the derivative \(f'(1)\) evaluated at the points \((1,1)\) and \((1,2)\) is \([8, -10]\). |
[8, -14] |
| deepseekr1 | To find the derivative \( f'(1) \) near the points \((1,1)\) and \((1,2)\) for the equation \( x^2 y^2 - 3y + 2x^3 = 0 \) where \( y = f(x) \), we use implicit differentiation. 1. **Differentiate the equation implicitly with respect to \( x \):** \[ \frac{d}{dx}(x^2 y^2 - 3y + 2x^3) = 0 \] Applying differentiation term by term: - For \( x^2 y^2 \): \( 2x y^2 + 2x^2 y y' \) - For \( -3y \): \( -3y' \) - For \( 2x^3 \): \( 6x^2 \) Combining these: \[ 2x y^2 + 2x^2 y y' - 3y' + 6x^2 = 0 \] 2. **Solve for \( y' \):** \[ y'(2x^2 y - 3) = - (2x y^2 + 6x^2) \] \[ y' = -\frac{2x y^2 + 6x^2}{2x^2 y - 3} \] 3. **Evaluate \( y' \) at the points \((1,1)\) and \((1,2)\):** - **At \((1,1)\):** \[ y' = -\frac{2(1)(1)^2 + 6(1)^2}{2(1)^2(1) - 3} = -\frac{8}{-1} = 8 \] - **At \((1,2)\):** \[ y' = -\frac{2(1)(2)^2 + 6(1)^2}{2(1)^2(2) - 3} = -\frac{14}{1} = -14 \] **Answer:** The derivatives \( f'(1) \) near the points \((1,1)\) and \((1,2)\) are \([8, -14]\). \boxed{[8, -14]} |
[8, -14] |
| gpto1 | To find \( f'(1) \) near the points \( (1,1) \) and \( (1,2) \), we will use implicit differentiation on the given equation \( x^2y^2 - 3y + 2x^3 = 0 \). **Step 1: Differentiate both sides with respect to \( x \).** Differentiate term by term, remembering that \( y \) is a function of \( x \) (i.e., \( y = f(x) \)): 1. \( \frac{d}{dx}(x^2y^2) = x^2 \cdot 2y \cdot y' + y^2 \cdot 2x = 2x^2yy' + 2xy^2 \) 2. \( \frac{d}{dx}(-3y) = -3y' \) 3. \( \frac{d}{dx}(2x^3) = 6x^2 \) Set the derivative equal to zero: \[ 2x^2yy' + 2xy^2 - 3y' + 6x^2 = 0 \] **Step 2: Solve for \( y' \).** Group \( y' \) terms and move others to the other side: \[ (2x^2y - 3)y' = - (2xy^2 + 6x^2) \] \[ y' = \frac{- (2xy^2 + 6x^2)}{2x^2y - 3} \] **Step 3: Evaluate \( y' \) at the given points.** **At \( (1,1) \):** \[ \begin{align*} y' &= \frac{- (2 \cdot 1 \cdot 1^2 + 6 \cdot 1^2)}{2 \cdot 1^2 \cdot 1 - 3} \\ &= \frac{- (2 + 6)}{2 - 3} \\ &= \frac{-8}{-1} \\ &= 8 \end{align*} \] **At \( (1,2) \):** \[ \begin{align*} y' &= \frac{- (2 \cdot 1 \cdot 2^2 + 6 \cdot 1^2)}{2 \cdot 1^2 \cdot 2 - 3} \\ &= \frac{- (8 + 6)}{4 - 3} \\ &= \frac{-14}{1} \\ &= -14 \end{align*} \] **Answer:** \[ [8,\ -14] \] |
[8, -14] |
| GPT-4o | To find the derivative \( f'(x) \) for the equation \( x^2 y^2 - 3y + 2x^3 = 0 \) when \( y = f(x) \), we will use implicit differentiation. We need to find \( f'(1) \) at the points (1,1) and (1,2). 1. **Implicit Differentiation:** Given: \[ x^2 y^2 - 3y + 2x^3 = 0 \] Differentiate both sides with respect to \( x \): \[ \frac{d}{dx} (x^2 y^2 - 3y + 2x^3) = \frac{d}{dx} 0 \] Using the product rule and the chain rule, we get: \[ 2x y^2 + x^2 \cdot 2y \frac{dy}{dx} - 3 \frac{dy}{dx} + 6x^2 = 0 \] Simplify: \[ 2x y^2 + 2x^2 y \frac{dy}{dx} - 3 \frac{dy}{dx} + 6x^2 = 0 \] Rearrange to solve for \( \frac{dy}{dx} \): \[ 2x y^2 + 6x^2 = 3 \frac{dy}{dx} - 2x^2 y \frac{dy}{dx} \] Factor out \( \frac{dy}{dx} \) on the right side: \[ 2x y^2 + 6x^2 = \left( 3 - 2x^2 y \right) \frac{dy}{dx} \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{2x y^2 + 6x^2}{3 - 2x^2 y} \] 2. **Evaluate \( f'(1) \) at the points (1,1) and (1,2):** - At the point (1,1): \[ \frac{dy}{dx} \bigg|_{(1,1)} = \frac{2 \cdot 1 \cdot 1^2 + 6 \cdot 1^2}{3 - 2 \cdot 1^2 \cdot 1} = \frac{2 + 6}{3 - 2} = \frac{8}{1} = 8 \] - At the point (1,2): \[ \frac{dy}{dx} \bigg|_{(1,2)} = \frac{2 \cdot 1 \cdot 2^2 + 6 \cdot 1^2}{3 - 2 \cdot 1^2 \cdot 2} = \frac{2 \cdot 4 + 6}{3 - 4} = \frac{8 + 6}{-1} = \frac{14}{-1} = -14 \] Therefore, the derivative \( f'(1) \) near the points (1,1) and (1,2) are: \[ f'(1) = \left[ 8, -14 \right] \] So the solution is: \[ \boxed{[8, -14]} \] |
[8, -14] |
| llama3.1-70B-Instruct | To find the derivative f'(x) of the given equation, we can use implicit differentiation. The equation is: x^2*y^2 - 3y + 2x^3 = 0 Differentiating both sides with respect to x, we get: 2xy^2 + x^2*2y*y' - 3y' + 6x^2 = 0 Now, we want to find the derivative f'(1) near the points (1,1) and (1,2). Near the point (1,1), we have x = 1 and y = 1. Substituting these values into the equation, we get: 2(1)(1)^2 + (1)^2*2(1)*y' - 3y' + 6(1)^2 = 0 Simplifying, we get: 2 + 2y' - 3y' + 6 = 0 Combine like terms: -y' + 8 = 0 y' = 8 So, f'(1) near the point (1,1) is 8. Near the point (1,2), we have x = 1 and y = 2. Substituting these values into the equation, we get: 2(1)(2)^2 + (1)^2*2(2)*y' - 3y' + 6(1)^2 = 0 Simplifying, we get: 8 + 8y' - 3y' + 6 = 0 Combine like terms: 5y' + 14 = 0 y' = -14/5 So, f'(1) near the point (1,2) is -14/5. Therefore, the list of answers is: [8, -14/5] |
[8, -14] |
| llama3.1-405B-Instruct | To find the derivative of y = f(x) near the points (1,1) and (1,2), we can use implicit differentiation. First, let's rewrite the equation as: x^2*y^2 - 3y + 2x^3 = 0 Now, we can differentiate both sides of the equation with respect to x, treating y as a function of x: d/dx (x^2*y^2) - d/dx (3y) + d/dx (2x^3) = 0 Using the product rule and the chain rule, we get: 2x*y^2 + x^2*2y*dy/dx - 3dy/dx + 6x^2 = 0 Now, we can solve for dy/dx: dy/dx = (3 - 2x*y^2 - 6x^2) / (2x^2*y + 3) Now, we can evaluate the derivative at the points (1,1) and (1,2): At (1,1): dy/dx = (3 - 2*1*1^2 - 6*1^2) / (2*1^2*1 + 3) = (3 - 2 - 6) / (2 + 3) = -5/5 = -1 At (1,2): dy/dx = (3 - 2*1*2^2 - 6*1^2) / (2*1^2*2 + 3) = (3 - 8 - 6) / (4 + 3) = -11/7 So, the derivative f'(1) near the point (1,1) is -1, and the derivative f'(1) near the point (1,2) is -11/7. The final answer is [-1, -11/7]. |
[8, -14] |
| DeepSeek-Coder-V2 | To solve the given problem, we need to find the derivative of \( y = f(x) \) at the points (1,1) and (1,2) for the equation \( x^2 y^2 - 3y + 2x^3 = 0 \). First, let's differentiate the equation implicitly with respect to \( x \): \[ x^2 y^2 - 3y + 2x^3 = 0 \] Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(x^2 y^2) - \frac{d}{dx}(3y) + \frac{d}{dx}(2x^3) = \frac{d}{dx}(0) \] Using the product rule and chain rule, we get: \[ \frac{d}{dx}(x^2 y^2) = 2x y^2 + x^2 \cdot 2y \cdot \frac{dy}{dx} = 2x y^2 + 2x^2 y \frac{dy}{dx} \] \[ \frac{d}{dx}(3y) = 3 \frac{dy}{dx} \] \[ \frac{d}{dx}(2x^3) = 6x^2 \] So, the equation becomes: \[ 2x y^2 + 2x^2 y \frac{dy}{dx} - 3 \frac{dy}{dx} + 6x^2 = 0 \] Rearrange to solve for \( \frac{dy}{dx} \): \[ 2x^2 y \frac{dy}{dx} - 3 \frac{dy}{dx} = -2x y^2 - 6x^2 \] \[ \frac{dy}{dx} (2x^2 y - 3) = -2x y^2 - 6x^2 \] \[ \frac{dy}{dx} = \frac{-2x y^2 - 6x^2}{2x^2 y - 3} \] Now, we need to find \( f'(1) \) at the points (1,1) and (1,2). 1. For the point (1,1): \[ \frac{dy}{dx} = \frac{-2(1)(1)^2 - 6(1)^2}{2(1)^2 (1) - 3} = \frac{-2 - 6}{2 - 3} = \frac{-8}{-1} = 8 \] 2. For the point (1,2): \[ \frac{dy}{dx} = \frac{-2(1)(2)^2 - 6(1)^2}{2(1)^2 (2) - 3} = \frac{-2(4) - 6}{2(2) - 3} = \frac{-8 - 6}{4 - 3} = \frac{-14}{1} = -14 \] So, the derivative \( f'(1) \) at the points (1,1) and (1,2) are 8 and -14, respectively. The answer in a list is: \[ \boxed{[8, -14]} \] |
[8, -14] |
| Qwen2-72B-Instruct | To find the derivative \(f'(x)\) of the given implicit function \(x^2y^2 - 3y + 2x^3 = 0\) with respect to \(x\), we need to perform implicit differentiation. After finding the derivative, we can then evaluate \(f'(1)\) at the points \((1,1)\) and \((1,2)\). Let's differentiate the equation with respect to \(x\): \[ \frac{d}{dx}(x^2y^2) - \frac{d}{dx}(3y) + \frac{d}{dx}(2x^3) = \frac{d}{dx}(0) \] Applying the product rule and chain rule where necessary: \[ 2xy^2 + x^2(2y\frac{dy}{dx}) - 3\frac{dy}{dx} + 6x^2 = 0 \] Rearranging for \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(2x^2y - 3) = -2xy^2 - 6x^2 \] Thus, \[ \frac{dy}{dx} = \frac{-2xy^2 - 6x^2}{2x^2y - 3} \] When \(x = 1\), the expression for \(\frac{dy}{dx}\) becomes: \[ \frac{dy}{dx} = \frac{-2y^2 - 6}{2y - 3} \] Now we can evaluate \(\frac{dy}{dx}\) at the points \((1,1)\) and \((1,2)\): 1. At \((1,1)\): \[ \frac{dy}{dx} = \frac{-2(1)^2 - 6}{2(1) - 3} = \frac{-8}{-1} = 8 \] 2. At \((1,2)\): \[ \frac{dy}{dx} = \frac{-2(2)^2 - 6}{2(2) - 3} = \frac{-10}{1} = -10 \] Therefore, the derivative \(f'(1)\) evaluated at the points \((1,1)\) and \((1,2)\) is \([8, -10]\). |
[8, -14] |
| baichuan4 | To find the derivative \(f'(x)\) of the function \(y = f(x)\) defined implicitly by the equation \(x^2y^2 - 3y + 2x^3 = 0\), we can use implicit differentiation. We differentiate both sides of the equation with respect to \(x\), while treating \(y\) as a function of \(x\). This gives us: \[2x(y)^2 + x^2(2yy') - 3y' + 6x^2 = 0\] Now, we need to solve for \(y'\), which is the derivative \(f'(x)\): \[2xy^2 + 2x^2yy' - 3y' + 6x^2 = 0\] Grouping terms with \(y'\): \[(2x^2y - 3)y' = -2xy^2 - 6x^2\] Solving for \(y'\): \[y' = \frac{-2xy^2 - 6x^2}{2x^2y - 3}\] Now, we can substitute the given points (1,1) and (1,2) into this derivative formula to find \(f'(1)\) at those points. For the point (1,1): \[f'(1) = \frac{-2(1)(1)^2 - 6(1)^2}{2(1)^2(1) - 3} = \frac{-2 - 6}{2 - 3} = \frac{-8}{-1} = 8\] For the point (1,2): \[f'(1) = \frac{-2(1)(2)^2 - 6(1)^2}{2(1)^2(2) - 3} = \frac{-8 - 6}{4 - 3} = \frac{-14}{1} = -14\] So, the list containing the derivatives at the points (1,1) and (1,2) would be: \[[8, -14]] |
[8, -14] |
| doubao | Differentiating the equation implicitly with respect to x: 2x*y^2 + 2x^2*y*y' - 3y' + 6x^2 = 0. Substituting (1,1): 2*1*1^2 + 2*1^2*1*y' - 3y' + 6*1^2 = 0 2 + 2y' - 3y' + 6 = 0 -y' + 8 = 0 y' = 8. Substituting (1,2): 2*1*2^2 + 2*1^2*2*y' - 3y' + 6*1^2 = 0 8 + 4y' - 3y' + 6 = 0 y' + 14 = 0 y' = -14. The answer is [8, -14]. |
[8, -14] |
| minimax | To find the derivative \( f'(x) \) of the implicitly defined function \( y = f(x) \) given by the equation \( x^2y^2 - 3y + 2x^3 = 0 \), we will use implicit differentiation. This involves differentiating both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \) (i.e., \( y = f(x) \)). Differentiating the equation implicitly with respect to \( x \), we get: \[ \frac{d}{dx}(x^2y^2) - \frac{d}{dx}(3y) + \frac{d}{dx}(2x^3) = 0 \] Applying the product rule to \( x^2y^2 \) and the chain rule to \( 2x^3 \), and remembering that \( \frac{dy}{dx} \) is \( f'(x) \), we have: \[ 2xy^2 + x^2(2y)\frac{dy}{dx} - 3\frac{dy}{dx} + 6x^2 = 0 \] Now we solve for \( \frac{dy}{dx} \), which is \( f'(x) \): \[ 2xy^2 + 2x^2y\frac{dy}{dx} - 3\frac{dy}{dx} + 6x^2 = 0 \] \[ (2x^2y - 3)\frac{dy}{dx} = -2xy^2 - 6x^2 \] \[ f'(x) = \frac{dy}{dx} = \frac{-2xy^2 - 6x^2}{2x^2y - 3} \] Now we need to evaluate \( f'(x) \) at \( x = 1 \) for the two points (1,1) and (1,2). For the point (1,1): \[ f'(1) = \frac{-2(1)(1)^2 - 6(1)^2}{2(1)^2(1) - 3} \] \[ f'(1) = \frac{-2 - 6}{2 - 3} \] \[ f'(1) = \frac{-8}{-1} \] \[ f'(1) = 8 \] For the point (1,2): \[ f'(1) = \frac{-2(1)(2)^2 - 6(1)^2}{2(1)^2(2) - 3} \] \[ f'(1) = \frac{-2(4) - 6}{2(2) - 3} \] \[ f'(1) = \frac{-8 - 6}{4 - 3} \] \[ f'(1) = \frac{-14}{1} \] \[ f'(1) = -14 \] So the list of the derivatives at the points (1,1) and (1,2) is: \[ [f'(1)_{(1,1)}, f'(1)_{(1,2)}] = [8, -14] \] |
[8, -14] |
| kimi | To find the derivative of \( f(x) \) at a point, we first need to express \( f(x) \) in terms of \( x \). Given the equation: \[ x^2y^2 - 3y + 2x^3 = 0 \] we can solve for \( y \) in terms of \( x \) to get \( f(x) \). Let's rearrange the equation to isolate \( y \): \[ y^2 = \frac{3y - 2x^3}{x^2} \] \[ y = f(x) = \sqrt{\frac{3 - 2x^3}{x^2}} \] Now, we need to find the derivative \( f'(x) \) using the chain rule. Let \( u = \frac{3 - 2x^3}{x^2} \), then \( f(x) = \sqrt{u} \). The derivative of \( f(x) \) with respect to \( x \) is: \[ f'(x) = \frac{1}{2\sqrt{u}} \cdot \frac{d}{dx}\left(\frac{3 - 2x^3}{x^2}\right) \] To find \( \frac{d}{dx}\left(\frac{3 - 2x^3}{x^2}\right) \), we use the quotient rule: \[ \frac{d}{dx}\left(\frac{3 - 2x^3}{x^2}\right) = \frac{(-6x^2)(x^2) - (3 - 2x^3)(2x)}{x^4} \] \[ = \frac{-6x^4 - 6x^4 + 4x^4}{x^4} \] \[ = \frac{-8x^4}{x^4} \] \[ = -8 \] Now we can substitute this back into our expression for \( f'(x) \): \[ f'(x) = \frac{1}{2\sqrt{\frac{3 - 2x^3}{x^2}}} \cdot (-8) \] \[ = \frac{-8}{2\sqrt{3 - 2x^3}} \cdot \frac{1}{x^2} \] \[ = \frac{-4}{x^2\sqrt{3 - 2x^3}} \] Now we need to evaluate \( f'(x) \) at \( x = 1 \) and \( x = 2 \): For \( x = 1 \): \[ f'(1) = \frac{-4}{1^2\sqrt{3 - 2(1)^3}} \] \[ = \frac{-4}{\sqrt{1}} \] \[ = -4 \] For \( x = 2 \): \[ f'(2) = \frac{-4}{2^2\sqrt{3 - 2(2)^3}} \] \[ = \frac{-4}{4\sqrt{3 - 16}} \] \[ = \frac{-4}{4\sqrt{-13}} \] Since the expression under the square root is negative, \( f'(2) \) is not real, which means \( f(x) \) is not differentiable at \( x = 2 \). Therefore, the derivative \( f'(1) \) near the point (1,1) is \( -4 \), and \( f'(2) \) does not exist because it would require taking the square root of a negative number. So the answer in a list is: \[ [-4, \text{undefined}] \] |
[8, -14] |
| wenxinyiyan | To solve this problem, we will use the implicit differentiation method to find the derivative of y with respect to x, treating y as a function of x. Then, we will find the values of this derivative at x = 1, when y = 1 and when y = 2. Given equation: $x^2y^2 - 3y + 2x^3 = 0$ Differentiating with respect to x, we get: $2xy^2 + 2x^2y\frac{dy}{dx} - 3\frac{dy}{dx} + 6x^2 = 0$ Now, solving for $\frac{dy}{dx}$, we get: $\frac{dy}{dx} = \frac{-2xy^2 - 6x^2}{2x^2y - 3}$ Now, we need to find $\frac{dy}{dx}$ at (1,1) and (1,2). For (1,1): $x = 1, y = 1$ $\frac{dy}{dx} = \frac{-2(1)(1)^2 - 6(1)^2}{2(1)^2(1) - 3} = \frac{-2 - 6}{2 - 3} = \frac{-8}{-1} = 8$ For (1,2): $x = 1, y = 2$ $\frac{dy}{dx} = \frac{-2(1)(2)^2 - 6(1)^2}{2(1)^2(2) - 3} = \frac{-8 - 6}{4 - 3} = \frac{-14}{1} = -14$ So, the values of the derivative near the points (1,1) and (1,2) are 8 and -14, respectively. Answer: [8, -14] |
[8, -14] |
| zhipu | To solve for the derivative \( f'(x) \) of the function \( y = f(x) \) described by the equation \( x^2y^2 - 3y + 2x^3 = 0 \), we will use implicit differentiation. Given the equation: \[ x^2y^2 - 3y + 2x^3 = 0 \] Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(x^2y^2) - \frac{d}{dx}(3y) + \frac{d}{dx}(2x^3) = 0 \] Applying the product rule and chain rule: \[ 2xy^2 + x^2(2y \frac{dy}{dx}) - 3 \frac{dy}{dx} + 6x^2 = 0 \] Simplify and group the terms involving \( \frac{dy}{dx} \): \[ 2xy^2 + 2x^2y \frac{dy}{dx} - 3 \frac{dy}{dx} + 6x^2 = 0 \] \[ 2x^2y \frac{dy}{dx} - 3 \frac{dy}{dx} = -2xy^2 - 6x^2 \] Factor out \( \frac{dy}{dx} \): \[ (2x^2y - 3) \frac{dy}{dx} = -2xy^2 - 6x^2 \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-2xy^2 - 6x^2}{2x^2y - 3} \] Now, we need to evaluate \( \frac{dy}{dx} \) at the points \( (1,1) \) and \( (1,2) \). 1. At the point \( (1,1) \): \[ \frac{dy}{dx} \bigg|_{(1,1)} = \frac{-2(1)(1)^2 - 6(1)^2}{2(1)^2(1) - 3} = \frac{-2 - 6}{2 - 3} = \frac{-8}{-1} = 8 \] 2. At the point \( (1,2) \): \[ \frac{dy}{dx} \bigg|_{(1,2)} = \frac{-2(1)(2)^2 - 6(1)^2}{2(1)^2(2) - 3} = \frac{-8 - 6}{4 - 3} = \frac{-14}{1} = -14 \] Therefore, the derivatives \( f'(1) \) near the points \( (1,1) \) and \( (1,2) \) are 8 and -14, respectively. Return the answer in a list: \[ [8, -14] \] |
[8, -14] |
| gpto1 | To find \( f'(1) \) near the points \( (1,1) \) and \( (1,2) \), we will use implicit differentiation on the given equation \( x^2y^2 - 3y + 2x^3 = 0 \). **Step 1: Differentiate both sides with respect to \( x \).** Differentiate term by term, remembering that \( y \) is a function of \( x \) (i.e., \( y = f(x) \)): 1. \( \frac{d}{dx}(x^2y^2) = x^2 \cdot 2y \cdot y' + y^2 \cdot 2x = 2x^2yy' + 2xy^2 \) 2. \( \frac{d}{dx}(-3y) = -3y' \) 3. \( \frac{d}{dx}(2x^3) = 6x^2 \) Set the derivative equal to zero: \[ 2x^2yy' + 2xy^2 - 3y' + 6x^2 = 0 \] **Step 2: Solve for \( y' \).** Group \( y' \) terms and move others to the other side: \[ (2x^2y - 3)y' = - (2xy^2 + 6x^2) \] \[ y' = \frac{- (2xy^2 + 6x^2)}{2x^2y - 3} \] **Step 3: Evaluate \( y' \) at the given points.** **At \( (1,1) \):** \[ \begin{align*} y' &= \frac{- (2 \cdot 1 \cdot 1^2 + 6 \cdot 1^2)}{2 \cdot 1^2 \cdot 1 - 3} \\ &= \frac{- (2 + 6)}{2 - 3} \\ &= \frac{-8}{-1} \\ &= 8 \end{align*} \] **At \( (1,2) \):** \[ \begin{align*} y' &= \frac{- (2 \cdot 1 \cdot 2^2 + 6 \cdot 1^2)}{2 \cdot 1^2 \cdot 2 - 3} \\ &= \frac{- (8 + 6)}{4 - 3} \\ &= \frac{-14}{1} \\ &= -14 \end{align*} \] **Answer:** \[ [8,\ -14] \] |
[8, -14] |
| deepseekv3 | To find the derivative \( f'(1) \) near the points \( (1,1) \) and \( (1,2) \), we will use implicit differentiation on the given equation: \[ x^2 y^2 - 3y + 2x^3 = 0 \] ### Step 1: Apply implicit differentiation Differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \): \[ \frac{d}{dx}(x^2 y^2) - \frac{d}{dx}(3y) + \frac{d}{dx}(2x^3) = \frac{d}{dx}(0) \] Using the product rule and chain rule, we get: \[ 2x y^2 + x^2 \cdot 2y \cdot y' - 3y' + 6x^2 = 0 \] Simplify the equation: \[ 2x y^2 + 2x^2 y y' - 3y' + 6x^2 = 0 \] ### Step 2: Solve for \( y' \) Collect the terms involving \( y' \): \[ 2x^2 y y' - 3y' = -2x y^2 - 6x^2 \] Factor out \( y' \): \[ y' (2x^2 y - 3) = -2x y^2 - 6x^2 \] Solve for \( y' \): \[ y' = \frac{-2x y^2 - 6x^2}{2x^2 y - 3} \] Simplify the expression: \[ y' = \frac{-2(x y^2 + 3x^2)}{2x^2 y - 3} \] \[ y' = \frac{-2x(y^2 + 3x)}{2x^2 y - 3} \] ### Step 3: Evaluate \( y'(1) \) at \( (1,1) \) and \( (1,2) \) #### At \( (1,1) \): Substitute \( x = 1 \) and \( y = 1 \) into the expression for \( y' \): \[ y' = \frac{-2(1)(1^2 + 3(1))}{2(1)^2 (1) - 3} \] \[ y' = \frac{-2(1 + 3)}{2 - 3} \] \[ y' = \frac{-2(4)}{-1} \] \[ y' = 8 \] #### At \( (1,2) \): Substitute \( x = 1 \) and \( y = 2 \) into the expression for \( y' \): \[ y' = \frac{-2(1)(2^2 + 3(1))}{2(1)^2 (2) - 3} \] \[ y' = \frac{-2(4 + 3)}{4 - 3} \] \[ y' = \frac{-2(7)}{1} \] \[ y' = -14 \] ### Final Answer: \[ [f'(1) \text{ at } (1,1), f'(1) \text{ at } (1,2)] = [8, -14] \] |
[8, -14] |