homework7: problem 1\n(2 points)\nfor the equation given below, evaluate $\\frac{dy}{dx}$ at the point $(-1…
homework7: problem 1\n(2 points)\nfor the equation given below, evaluate $\\frac{dy}{dx}$ at the point $(-1, -2)$.\n$4y^{3}+y^{2}-4x^{2}=-32$\n$\\frac{dy}{dx}$ at $(-1, -2)=$\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 19, 2025, 8:36:21 pm cdt\nwebwork © 1996 - 2024 | theme: math4_ttu | ww_version: 2.19 | pg_version 2.19\nthe webwork project
Answer
Explanation:
Step1: Differentiate both sides with respect to (x)
Differentiate (4y^{3}+y^{2}-4x^{2}=-32) term - by - term. Using the chain rule ((u^{n})^\prime = nu^{n - 1}u^\prime) (where (u = y) and (u^\prime=\frac{dy}{dx})) and ((x^{n})^\prime=nx^{n - 1}): (\frac{d}{dx}(4y^{3})+\frac{d}{dx}(y^{2})-\frac{d}{dx}(4x^{2})=\frac{d}{dx}(-32)) (4\times3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-4\times2x = 0) (12y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-8x = 0)
Step2: Solve for (\frac{dy}{dx})
Factor out (\frac{dy}{dx}): (\frac{dy}{dx}(12y^{2}+2y)=8x) Then (\frac{dy}{dx}=\frac{8x}{12y^{2}+2y}=\frac{4x}{6y^{2}+y})
Step3: Substitute (x=-1) and (y = - 2)
(\frac{dy}{dx}\mid_{x=-1,y = - 2}=\frac{4\times(-1)}{6\times(-2)^{2}+(-2)}) First, calculate the denominator: (6\times(-2)^{2}+(-2)=6\times4-2=24 - 2=22) Then (\frac{4\times(-1)}{22}=-\frac{2}{11})
Answer:
(-\frac{2}{11})