# derivative trigonometry

• Sep 11th 2008, 08:51 PM
jvignacio
derivative trigonometry
$\displaystyle f(x) = tanxsinx$

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

this correcT?
• Sep 11th 2008, 08:58 PM
11rdc11
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 PM
jvignacio
Quote:

Originally Posted by 11rdc11
Yep

can we simplify this ?
if so how would u do it?
• Sep 11th 2008, 09:02 PM
Chris L T521
Quote:

Originally Posted by jvignacio
$\displaystyle f(x) = tanxsinx$

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

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
• Sep 11th 2008, 09:07 PM
jvignacio
Quote:

Originally Posted by Chris 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

what if the possible answers were:

$\displaystyle sinx - sinxsec^2x - sinx(sec^2x+1) - secx - none of these$

what would you go for?
• Sep 11th 2008, 09:16 PM
11rdc11
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 PM
jvignacio
Quote:

Originally Posted by 11rdc11
I did this kind of fast but I'm pretty sure it the 3rd option

yeah thats what i put but i wasnt sure why we got rid of the tanx and cosx