If `f(x)=x^n` where `x in R` , then differentiation of `x^n` with respect to `x` `is x^(n-1)dot`

Explanation

Given,`f(x)=x^n`

`g(x)=x sin x^(n-1)`

so,

`((d(x^n))/(d(x sin x^(n-1))))`

=`(d/dx(f(x)))/(d/dx(g(x))`

now,`f'(x)=(d/dx(x^n))=nx^(n-1)`

`g'(x)=d/dx(x sin x^(n-1))`

=`sinx^(n-1)+ cosx^(n-1)(n-1)x^(n-1)`

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