Question #09e0a

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Answer 1 · 2017-03-01T03:11:52

$dy/dx = 4tan^3(x)sec^2(x) + 5x^4sec^2(x^5)$

Use the chain rule on each term.

First term: $y_1 = tan^4(x)$

Let $u = tan(x)$ , then $y = u^4, dy/(du)=4u^3, and (du)/dx = sec^2(x)$

$dy_1/dx = dy/(du)(du)/dx$

$dy_1/dx = 4u^3sec^2(x)$

$dy_1/dx = 4tan^3(x)sec^2(x)$

Second term: $y_2 = tan(x^5)$

Let $u = x^2$ , then $y = tan(u), dy/(du)=sec^2(u), and (du)/dx = 5x^4$

$dy_2/dx = dy/(du)(du)/dx$

$dy_2/dx = sec^2(u)5x^4$

$dy_2/dx = 5x^4sec^2(x^5)$

Put both terms back into the expression:

$dy/dx = dy_1/dx + dy_2/dx$

$dy/dx = 4tan^3(x)sec^2(x) + 5x^4sec^2(x^5)$