First derivative of tan (xsinx)

Question

zepdrix · Answer

Start with chain rule:$$\Large\bf\sf \left[\tan(x \sin x)\right]'\quad=\quad \sec^2(x \sin x)\cdot \color{royalblue}{(x \sin x)'}$$Derivative of the outer function tangent, gave us secant squared.
Chain rule tells us to multiply by the derivative of the inner function.
So we have to take the derivative of this blue part.
Make sense so far?

anonymous · Answer

yes yes. Thank you =)

zepdrix · Answer

Understand what to do next? :o
Hint hint `product rule`

anonymous · Answer

okay. wait. i'll try..

anonymous · Answer

sec^2 (xsinx) * (xcosx)+(sinx)

zepdrix · Answer

Ok very good!
Let's just make sure we put brackets around our product rule to show that the entire thing is multiplying the secant.

sec^2 (xsinx) * [(xcosx)+(sinx)]

zepdrix · Answer

Or maybe like this so it's a little easier to read :)

sec^2 (xsinx) * [xcosx+sinx]
But whatev :) Looks like you've got it

anonymous · Answer

aw.. this kills me. my answer was tanx (xcox+sinx) but thanks =))))