Implicit Differentiation, Plugging in Values

Question

Suppose that $(g(x))^2+16x=x^2g(x)+1$ and that $g(4)=7$ Find $g'(4)$

So for the derivative, I got $(2xg(x)-16)/(2g(x)-x^2)$

After plugging in the values, I got $\frac{10}3$ for my answer. Could someone let me know what I am doing wrong?

Answer 1

Let's do the full working.

$$(g(x))^2+16x=x^2g(x)+1$$

differentiating both sides w.r.t. $x$ yields

$$2g'(x)g(x)+16=2xg(x)+x^2g'(x)$$

rearranging to find an expression for $g'(x)$:

\begin{align*} 2g'(x)g(x)+16&=2xg(x)+x^2g'(x) \\ 2g'(x)g(x)-x^2g'(x)&=2xg(x)-16 \\ g'(x)(2g(x)-x^2) &= 2xg(x)-16 \\ g'(x) &= \frac{2xg(x)-16}{2g(x)-x^2} \end{align*}

so your expression for the derivative is indeed correct.

Therefore:

$$g'(4) = \frac{2\cdot4\cdot7-16}{2\cdot7-4^2}=\frac{40}{-2}=-20$$

not quite sure how you got $\frac{10}{3}$.