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 :

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

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

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

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