use implicit differentiation to find y’ and then evaluate y’ at (2,0). 32e^y = x^5 + y^5 y’ = y’|(2,0) =…

Explanation:

Step1: Differentiate both sides

Differentiate $32e^{y}=x^{5}+y^{5}$ with respect to $x$. The derivative of $32e^{y}$ with respect to $x$ using the chain - rule is $32e^{y}y'$. The derivative of $x^{5}$ with respect to $x$ is $5x^{4}$, and the derivative of $y^{5}$ with respect to $x$ using the chain - rule is $5y^{4}y'$. So we have $32e^{y}y'=5x^{4}+5y^{4}y'$.

Step2: Isolate $y'$

Move all terms with $y'$ to one side: $32e^{y}y' - 5y^{4}y'=5x^{4}$. Factor out $y'$: $y'(32e^{y}-5y^{4}) = 5x^{4}$. Then $y'=\frac{5x^{4}}{32e^{y}-5y^{4}}$.

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

Substitute $x = 2$ and $y = 0$ into $y'$. $y'\big|_{(2,0)}=\frac{5\times2^{4}}{32e^{0}-5\times0^{4}}=\frac{5\times16}{32\times1 - 0}=\frac{80}{32}=\frac{5}{2}$.

Answer:

$y'=\frac{5x^{4}}{32e^{y}-5y^{4}}$ $y'\big|_{(2,0)}=\frac{5}{2}$