In the given expression
[tex]x^5+32y^3=\ln y[/tex]To differentiate it with respect to it, we will writ y' for every term of y
The differentiation of
[tex]x^5=5x^{5-1}=5x^4[/tex]The differentiation of
[tex]32y^3=32(3)y^{3-1}y^{\prime}=96y^2y^{\prime}[/tex]The differentiation of
[tex]\ln y=\frac{1}{y}(y^{\prime})=y^{-1}y^{\prime}[/tex]Now let us write all of them in one line
[tex]5x^4+96y^2y^{\prime}=y^{-1}y^{\prime}[/tex]Put the terms of y' on one side and the other term on the other side
[tex]96y^2y^{\prime}-y^{-1}y^{\prime}=-5x^4[/tex]Take y' as a common factor
[tex]y^{\prime}(96y^2-y^{-1})=-5x^4[/tex]Divide both sides by the bracket
[tex]y^{\prime}=\frac{-5x^4}{(96y^2-y^{-1})}[/tex]Since the given point is (-2, 1), then
x = -2 and y = 1
Substitute them to find y'
[tex]\begin{gathered} y^{\prime}=\frac{-5(-2)^4}{(96\lbrack1\rbrack^2-\lbrack1\rbrack^{-1})}=\frac{-5(16)}{(96-1)}=\frac{-80}{95} \\ y^{\prime}=-\frac{16}{19} \end{gathered}[/tex]The value of y' is -16/19
