If $4x^2+5x+xy=4$ and $y(4)=-20$, find $y'(4)$ by implicit differentiation

Question

If $4x^2+5x+xy=4$ and $y(4)=-20$, find $y'(4)$ by implicit differentiation.

I implicitly differentiated the equation, but I don't see how I can use $y(4)=-20$ to my advantage.

Answer 1

From $4x^2+5x+xy=4$ you get

$$8x+5+y(x)+xy'(x)=0.$$

If $x=4$ we have

$$32+5+y(4)+4y'(4)=0.$$

Can you finish now?

Answer 2

Hint:

$$8x+ 5 + y(x) + xy' = 0 \Rightarrow y'(x) = \frac{-(y(x) + 8x + 5)}{x}$$