use implicit differentiation to find y and then evaluate y at (-2,0). 16e^y = x^4 - y^5\n\n y = \n…

Explanation:

Step1: Differentiate both sides

Differentiate $16e^{y}=x^{4}-y^{5}$ with respect to $x$. The derivative of $16e^{y}$ using the chain - rule is $16e^{y}y'$, the derivative of $x^{4}$ is $4x^{3}$, and the derivative of $-y^{5}$ using the chain - rule is $-5y^{4}y'$. So we have $16e^{y}y'=4x^{3}-5y^{4}y'$.

Step2: Isolate $y'$

Move all terms with $y'$ to one side: $16e^{y}y'+5y^{4}y' = 4x^{3}$. Factor out $y'$: $y'(16e^{y}+5y^{4})=4x^{3}$. Then $y'=\frac{4x^{3}}{16e^{y}+5y^{4}}$.

Step3: Evaluate $y'$ at $(-2,0)$

Substitute $x = - 2$ and $y = 0$ into $y'$. We know that $e^{0}=1$. So $y'\big|_{(-2,0)}=\frac{4(-2)^{3}}{16e^{0}+5(0)^{4}}=\frac{4\times(-8)}{16\times1 + 0}=\frac{-32}{16}=-2$.

Answer:

$y'=\frac{4x^{3}}{16e^{y}+5y^{4}}$ $y'\big|_{(-2,0)}=-2$