Question

I need help with this question

Answer 1

In the given expression

$x^5+32y^3=\ln y$

To differentiate it with respect to it, we will writ y' for every term of y

The differentiation of

$x^5=5x^{5-1}=5x^4$

The differentiation of

$32y^3=32(3)y^{3-1}y^{\prime}=96y^2y^{\prime}$

The differentiation of

$\ln y=\frac{1}{y}(y^{\prime})=y^{-1}y^{\prime}$

Now let us write all of them in one line

$5x^4+96y^2y^{\prime}=y^{-1}y^{\prime}$

Put the terms of y' on one side and the other term on the other side

$96y^2y^{\prime}-y^{-1}y^{\prime}=-5x^4$

Take y' as a common factor

$y^{\prime}(96y^2-y^{-1})=-5x^4$

Divide both sides by the bracket

$y^{\prime}=\frac{-5x^4}{(96y^2-y^{-1})}$

Since the given point is (-2, 1), then

x = -2 and y = 1

Substitute them to find y'

$\begin{gathered} y^{\prime}=\frac{-5(-2)^4}{(96\lbrack1\rbrack^2-\lbrack1\rbrack^{-1})}=\frac{-5(16)}{(96-1)}=\frac{-80}{95} \\ y^{\prime}=-\frac{16}{19} \end{gathered}$

The value of y' is -16/19