dy/ dx of y=sin^3xtan4x

Question

anonymous · Answer

you can use the product rule

amistre64 · Answer

good thing it aint the integral

anonymous · Answer

Product and chain rule.

anonymous · Answer

@amistre64:shouldn't by parts work then ?

amistre64 · Answer

it might, but i dunno

electrochika · Answer

I don't know how to multiply it in.

amistre64 · Answer

aint got to multiply it in; product rule is:

[fg]' = f'g + fg'

anonymous · Answer

just find the derivaitve of sin^3x and tan4x seperatly, then you can put them together

electrochika · Answer

the first times the derivative of the second plus the second times the derivative of the first?

amistre64 · Answer

yes

amistre64 · Answer

or if you have more than 2 it just a sum that allows each one to get derived seperately

amistre64 · Answer

[fgh]' = f'gh + fg'h + fgh'

electrochika · Answer

i set it up in the product rule but now i'm stuck

amistre64 · Answer

what you got?

anonymous · Answer

$$ 4 \text{Sec}[4 x]^2 \text{Sin}[x]^3+3 \text{Cos}[x] \text{Sin}[x]^2 \text{Tan}[4 x] $$


I am feeling lazy :P

electrochika · Answer

sin^3x((sec^2(4x)) + tan4x(cos^3x) is that even right?

anonymous · Answer

I gave you the correct answer

electrochika · Answer

I need to show steps

anonymous · Answer

OHHHH.. then follow the instructions of amistre...

anonymous · Answer

what instructions? use 
$$(fg)'=f'g+g'f$$ with 
$$f(x)=\sin^3(x),f'(x)=3\sin^2(x)\cos(x),g(x)=\tan(4x), g'(x)=4\sec^2(x)$$ and put it together