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

The value of y' is -16/19