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Question
question. find the equation of the tangent line at the point (-3,2) to the curve defined implicitly below. (x^{4}+y^{3}=89). give your answer in the form (y = mx + b). provide your answer below:
Step1: Differentiate implicitly
Differentiate $x^{4}+y^{3}=89$ with respect to $x$: $4x^{3}+3y^{2}y' = 0$.
Step2: Solve for $y'$
$y'=-\frac{4x^{3}}{3y^{2}}$.
Step3: Find slope at $(-3,2)$
Substitute $x = - 3$ and $y = 2$ into $y'$: $m=-\frac{4(-3)^{3}}{3(2)^{2}}=9$.
Step4: Find $b$
Use $y=mx + b$ with $x=-3,y = 2,m = 9$. So $2=9(-3)+b$, then $b = 29$.
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$y=9x + 29$