find g'(2) if g(x)=x^3h(cos(pi*x)), where h(1) = 2 h'(1)=-2
im really stuck on this one and frankly don't even know where to start
thanks so much
The standard method to find the derivative of functions of type $\displaystyle f^g$ where $\displaystyle f$ and $\displaystyle g$ are functions of $\displaystyle x$ is by taking logarithm on both sides and proceeding.
On the right side you have this very occurrence. Here, $\displaystyle f = x$ while $\displaystyle g = 3h(\cos \pi\cdot x)$.
Let's take log both sides and proceed...
$\displaystyle \log g(x)=3h(\cos \pi\cdot x) + \log(x)$.
Differentiating w.r.t. $\displaystyle x$ throughout...
$\displaystyle 1/g(x) \cdot g(x)^\prime=3h^\prime(\cos \pi\cdot x) \cdot -\pi\sin(\pi\cdot x)+1/x\quad(2)$
In your original equation if we put x=2,
$\displaystyle g(2)=2^{3h(cos(\pi \cdot 2))} =2^{3h(\cos 0)}=2^{3h(1)}=2^{3\times-2}$ which is not $\displaystyle 0$ so we can put $\displaystyle x=2$ in the eqn 2 abv.
Which if we do, the right side of eqn 2 takes the value 1/2.
So, we have, $\displaystyle g^\prime(2)/g(2)=1/2$.
So, you can find $\displaystyle g^\prime(2)$, now
I thought the problem was $\displaystyle g(x)=x^{3\cdot h(\cos \pi\cdot x)}$ while it is actually:
$\displaystyle g(x)=x^3\cdot h(\cos \pi\cdot x)\quad(*)$
Better if you had used Latex or I had interpreted you better.
Anyway,
you can simply differentiate that equation w.r.t. x throughout and put x=2 in the equation that follows.
Let's do that...
$\displaystyle d(g(x))/dx=d(x^3)/dx\cdot h(\cos \pi\cdot x) \,+\, x^3\cdot d(h(\cos \pi\cdot x))/dx$
$\displaystyle or, g^\prime(x)=3x^2\cdot h(\cos \pi\cdot x)\,+\,x^3 h^\prime(\cos \pi\cdot x)\cdot -\pi\cdot\sin (\pi\cdot x)$
In the equation and the equation $\displaystyle (*)$ above we put $\displaystyle x=2$ and get $\displaystyle g^\prime(2)$
OK, now?
thank you very much, it was actually a simple problem :S just that h(1)=2 confused me but now i understand.. thank you
oh by the way my final answer was g'(2)=8 is that correct?
also sorry about not using latex when i have some more time ill figure it out
thanks again