# Thread: Deriviative of function #2

1. ## Deriviative of function #2

i)Find h'(2) given that h(x)=f(g(x)), f(u)=u^2-1, g(2)=3, g'(2)=-1

ii) let y=f(x^2+3x-5). find dy/dx when x=1, given that f'(-1)=2

2. $h(x)=f(g(x))$

$f(u)=u^2-1$

$f(g(x))=(g(x))^2-1$

Just use the chain rule:

$h'(x) = 2g(x)g'(x)$

You can now just substitute to get $h'(2)$

3. $y(x) = f(x^{2}+3x-5)$

$y'(x) = f'(x^{2}+3x-5) \frac{d}{dx} (x^{2}+3x-5) = f'(x^{2}+3x-5)(2x+3)$

$y'(1) = f'(-1)(5) = 10$

4. $y=f(g(x)$

Let $g(x)=x^2+3x-5$

$\frac{dy}{dx}=f'((g(x))g'(x)$

$=f'(g(x))[2x+3]$

When $x=1, g(1)=-1$, so we have $\frac{dy}{dx}=5f'(-1)$

5. Originally Posted by adkinsjr
$y=f(g(x)$

Let $g(x)=x^2+3x-5$

$\frac{dy}{dx}=f'((g(x))g'(x)$

$=f'(g(x))[2x+3]$

When $x=1, g(1)=-1$, so we have $\frac{dy}{dx}=5f'(-1)$
where did the 5 come from in the last line

6. Originally Posted by skeske1234
where did the 5 come from in the last line
The 5 is the derivative $g'(x)=2x+3$ at $x=1$

7. Originally Posted by adkinsjr
$h(x)=f(g(x))$

$f(u)=u^2-1$

$f(g(x))=(g(x))^2-1$

Just use the chain rule:

$h'(x) = 2g(x)g'(x)$

You can now just substitute to get $h'(2)$

not sure what you mean.... i sorry don't follow that.

8. Originally Posted by skeske1234
not sure what you mean.... i sorry don't follow that.
We have the function $f(u)=u^2-1$ if we set $u=g(x)$ we get the function $h(x)=f(g(x))=(g(x))^2-1$ (notice how I substituted $u=g(x)$). We use the chain rule to differentiate:

$h'(x)=f'(g(x))g'(x)$.

$=2g(x)g'(x)$.

Do you understand how I got $h'(x)=2g(x)g'(x)$?

If so, then we have

$h'(2)=2g(2)g'(2)$

$h'(2)=2(3)(-1)=-6$