$\displaystyle
f(x) = tanxsinx
$
$\displaystyle
f '(x) = (sec^2x)(sinx) + (tanx)(cosx)
$
this correcT?
Yes, but it can be simplified further:
Note that $\displaystyle \sec^2x\sin x=\frac{1}{\cos^2x}\sin x=\tan x \sec x$ and $\displaystyle \tan x\cos x=\frac{\sin x}{\cos x}\cos x=\sin x$
So we now see that $\displaystyle f '(x) = (\sec^2x)(\sin x) + (\tan x)(\cos x)=\sec x\tan x+\sin x$
--Chris
I did this kind of fast but I'm pretty sure it the 3rd option
$\displaystyle
f '(x) =\sec x\tan x+\sin x
$
$\displaystyle \frac{sinx}{cos^2x} + sinx$
$\displaystyle \frac{sinx+sinxcos^2x}{cos^2x}$
$\displaystyle \frac{sinx(1+cos^2x)}{cos^2x}$
$\displaystyle (sinx)(sec^2x + 1)$