Question #09e0a

1 Answer
Mar 1, 2017

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

Explanation:

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)#