How do you differentiate # y =x /sec ^2x^3# using the chain rule?

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Answer 1 · 2018-07-04T04:30:29

#(dy)/(dx)=cosx^3[cosx^3-6x^3sinx^3]#

Here,

#y=x/sec^2(x^3)#

#=>y=x*cos^2(x^3)#

Diff.w.r.t. #x# , using Product rule and Chain Rule:

#(dy)/(dx)=x*d/(dx)(cos^2x^3)+cos^2x^3*d/(dx)(x)#

#=>(dy)/(dx)=x*2cosx^3(-sinx^3)d/(dx)(x^3)+cos^2x^3*1#

#=>(dy)/(dx)=-2xsinx^3cosx^3*3x^2+cos^2x^3#

#=>(dy)/(dx)=-6x^3sinx^3cosx^3+cos^2x^3#

#=>(dy)/(dx)=cos^2x^3-6x^3sinx^3cosx^3...to(1)#

#=>(dy)/(dx)=cosx^3[cosx^3-6x^3sinx^3]#

Note :

From #(1)# we can simplify in the form :

#(dy)/(dx)=cos^2x^3-6x^3sinx^3cosx^3#

#=>(dy)/(dx)=cos^2x^3-3x^3{2sinx^3cosx^3}#

#=>(dy)/(dx)=cos^2x^3-3x^3*sin2(x^3)#