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sleet_krkn [62]
3 years ago
8

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

Mathematics
2 answers:
dmitriy555 [2]3 years ago
8 0
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)}
strojnjashka [21]3 years ago
6 0
<span>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</span>
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If you need a more mathematical proof, you could do it this way.

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