use implicit differentiation to find y and then evaluate y at the point (2,3). y - 4x^3 + 5 = 0 y = y|_(2,3)…

use implicit differentiation to find y and then evaluate y at the point (2,3). y - 4x^3 + 5 = 0 y = y|_(2,3) = (simplify your answer.)
Answer
Explanation:
Step1: Differentiate each term
Differentiate $y - 4x^{3}+5 = 0$ with respect to $x$. The derivative of $y$ with respect to $x$ is $y'$, the derivative of $-4x^{3}$ is $-12x^{2}$ and the derivative of the constant 5 is 0. So we get $y'-12x^{2}=0$.
Step2: Solve for $y'$
Add $12x^{2}$ to both sides of the equation $y'-12x^{2}=0$. We have $y' = 12x^{2}$.
Step3: Evaluate $y'$ at the point $(2,3)$
Substitute $x = 2$ into $y'=12x^{2}$. Then $y'\big|_{(2,3)}=12\times2^{2}=12\times 4 = 48$.
Answer:
$y' = 12x^{2}$ $y'\big|_{(2,3)}=48$