# derivatives

#### steezin

Let y = g(h(x)) where h(x) = x^2/x+2. If g'(9/5) = -2, find dy/dx when x = 3

#### skeeter

MHF Helper
Let y = g(h(x)) where h(x) = x^2/x+2. If g'(9/5) = -2, find dy/dx when x = 3
chain rule ...

$$\displaystyle \frac{dy}{dx} = g'[h(x)] \cdot h'(x)$$

note that $$\displaystyle h(3) = \frac{9}{5}$$

finish it.

steezin

#### steezin

I know how to derive the inner function but I don't know how to find the derivative of g(h(x))?.

#### skeeter

MHF Helper
$$\displaystyle \frac{dy}{dx} = g'[h(x)] \cdot h'(x)$$

$$\displaystyle \frac{dy}{dx} = g'\left(\frac{x^2}{x+2}\right) \cdot \frac{(x+2)(2x)-(x^2)(1)}{(x+2)^2}$$

now evaluate for $$\displaystyle x = 3$$

steezin

#### steezin

So I would get dy/dx = x^4 + 4x^3/(x + 2)^3? and sub in 3 for x. The g' is confusing me, does it just disappear or something?

#### skeeter

MHF Helper
$$\displaystyle \frac{dy}{dx} = g'\left(\frac{x^2}{x+2}\right) \cdot \frac{(x+2)(2x)-(x^2)(1)}{(x+2)^2}$$

sub in 3 ...

$$\displaystyle \frac{dy}{dx} = \textcolor{red}{g'\left(\frac{3^2}{3+2}\right)} \cdot \frac{(3+2)(2\cdot 3)-(3^2)(1)}{(3+2)^2}$$

now, from your original post ...

If g'(9/5) = -2, find dy/dx when x = 3

steezin

ohhh, ok thanks!