Question

Derivative of y = cos(sqrt(sin(tan pi x))) ?

Answer 1

Hello,

$(cos( \sqrt{sin(tg( \pi x))} )'=- \dfrac{x sin( \sqrt{sin(tg( \pi x))} *cos(tg( \pi x)* \pi }{2 \sqrt{sin(tg( \pi x))} cos^2( \pi x)}$

Answer 2

Did your teacher really give you this problem? What a pain in the neck (hint the word I was gonna put here starts with an a). But it is good practice of... Chain rule: d/du(u) = 1 dx(tan(u)) = sec^2(u) * du/dx d/dx of sin(u) = cos(u) * du/dx d/dx(sqrt(u)) = 1/2u^(-1/2) * du/dx d/dx(cos(u)) = -sin(u) * du/dx Therefore: \[\frac{d}{dx}(\cos(\sqrt(\sin(\tan (\pi x)))))) = -\sin(\sqrt(\sin(\tan (\pi x)))) * \frac{1}2[\sin(\tan (\pi x)))]^{-1/2}\] \[* \cos(\tan (\pi x)) * \sec^2(\pi x) * \pi\] I had to break it into two lines because it was so long that it wouldn't fit on one line lol