How do you differentiate #f(x)=tan((1/cos(7x))^2)# using the chain rule?

1 Answer
Nov 10, 2015

Answer:

#dy/dx = 14sec^2(sec^2(7x))tan(7x)sec^2(7x)#

Explanation:

Okay, so, we have

#y = tan(sec^2(7x))#

Call #sec^2(7x) = u#

#dy/dx = d/(du)tan(u)(du)/dx = sec^2(u)(d)/dxsec^2(7x)#

Call #sec(7x) = v#

#dy/dx = sec^2(u)d/(dv)v^2(dv)/dx = 2v*sec^2(u)(d)/dx(sec(7x))#

Call #7x = w#

#dy/dx = 2v*sec^2(u)d/(dw)sec(w)(dw)/dx#
#dy/dx = 2v*sec^2(u)*tan(w)sec(w)d/dx(7x)#
#dy/dx = 14v*sec^2(u)*tan(w)sec(w)#

Substituting #w# back to #x#

#dy/dx = 14v*sec^2(u)*tan(7x)sec(7x)#

Substituting #v# back to #x#

#dy/dx = 14*sec^2(u)*tan(7x)sec^2(7x)#

Substituing #u# back to #x#

#dy/dx = 14sec^2(sec^2(7x))tan(7x)sec^2(7x)#