1. ## derivative trigonometry

$

f(x) = tanxsinx

$

$

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

$

this correcT?

2. Yep

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

You could simplify it a little more

$tan(x)(secx+cosx)$

3. Originally Posted by 11rdc11
Yep
can we simplify this ?
if so how would u do it?

4. Originally Posted by jvignacio
$

f(x) = tanxsinx

$

$

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

$

this correcT?
Yes, but it can be simplified further:

Note that $\sec^2x\sin x=\frac{1}{\cos^2x}\sin x=\tan x \sec x$ and $\tan x\cos x=\frac{\sin x}{\cos x}\cos x=\sin x$

So we now see that $f '(x) = (\sec^2x)(\sin x) + (\tan x)(\cos x)=\sec x\tan x+\sin x$

--Chris

5. Originally Posted by Chris L T521
Yes, but it can be simplified further:

Note that $\sec^2x\sin x=\frac{1}{\cos^2x}\sin x=\tan x \sec x$ and $\tan x\cos x=\frac{\sin x}{\cos x}\cos x=\sin x$

So we now see that $f '(x) = (\sec^2x)(\sin x) + (\tan x)(\cos x)=\sec x\tan x+\sin x$

--Chris
what if the possible answers were:

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

what would you go for?

6. I did this kind of fast but I'm pretty sure it the 3rd option

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

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

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

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

$(sinx)(sec^2x + 1)$

7. 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