$\displaystyle

f(x) = tanxsinx

$

$\displaystyle

f '(x) = (sec^2x)(sinx) + (tanx)(cosx)

$

this correcT?

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- Sep 11th 2008, 08:51 PMjvignacioderivative trigonometry
$\displaystyle

f(x) = tanxsinx

$

$\displaystyle

f '(x) = (sec^2x)(sinx) + (tanx)(cosx)

$

this correcT? - Sep 11th 2008, 08:58 PM11rdc11
Yep

$\displaystyle

f '(x) = (sec^2x)(sinx) + (tanx)(cosx)

$

You could simplify it a little more

$\displaystyle tan(x)(secx+cosx)$ - Sep 11th 2008, 09:00 PMjvignacio
- Sep 11th 2008, 09:02 PMChris L T521
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 - Sep 11th 2008, 09:07 PMjvignacio
- Sep 11th 2008, 09:16 PM11rdc11
I did this kind of fast but I'm pretty sure it the 3rd option

$\displaystyle

f '(x) =\sec x\tan x+\sin x

$

$\displaystyle \frac{sinx}{cos^2x} + sinx$

$\displaystyle \frac{sinx+sinxcos^2x}{cos^2x}$

$\displaystyle \frac{sinx(1+cos^2x)}{cos^2x}$

$\displaystyle (sinx)(sec^2x + 1)$ - Sep 11th 2008, 09:27 PMjvignacio