Let f(x)= sqrt(cos(e^((x^2)*cosx))) Find f'(x)



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cosinusix's profile pic

Posted on (Answer #1)

`f'(x)=(1/2)(cos(e^(x^2cos x))')/sqrt(cos(e^((x^2)*cosx)))`


`cos(e^(x^2cosx))'=-(e^(x^2cos x))'sin(e^(x^2cosx))`

`(e^(x^2cosx))'=(2xcos x-x^2sin x)e^(x^2 cos x)`


answer` f'(x)=-(1/2)(2xcos x-x^2sin x)e^(x^2 cos x)sin(e^(x^2cosx))/sqrt(cos(e^((x^2)*cosx)))`



sciencesolve's profile pic

Posted on (Answer #1)

You need to notice that the function is composed, hence you need to use chain rule to differentiate with respect to x such that:

`f'(x) = (sqrt(cos(e^((x^2)*cosx))))'`

`f'(x) = ((cos(e^((x^2)*cosx))'))/(2sqrt(cos(e^((x^2)*cosx))))`

`f'(x) = ((e^((x^2)*cosx))'((-sin(e^((x^2)*cosx))')))/(2sqrt(cos(e^((x^2)*cosx))))`

`f'(x) = ((x^2)*cosx)'(e^((x^2)*cosx))(-sin(e^((x^2)*cosx)))/(2sqrt(cos(e^((x^2)*cosx))))`

You need to use product rule to find derivative (x^2)*cosx such that:

`((x^2)*cosx)' = (x^2)'*cosx + (x^2)*(cosx)'`

`((x^2)*cosx)' = 2xcosx + (x^2)*(-sin x)`

`((x^2)*cosx)' = 2xcosx - (x^2)*(sin x)`

You need to substitute `2xcosx - (x^2)*(sin x)`  for `((x^2)*cosx)' ` such that:

`f'(x) = ((2xcosx - (x^2)*(sin x))(e^((x^2)*cosx))(-sin(e^((x^2)*cosx))))/(2sqrt(cos(e^((x^2)*cosx))))`

Hence, evaluating derivative of function yields `f'(x) = ((2xcosx - (x^2)*(sin x))(e^((x^2)*cosx))(-sin(e^((x^2)*cosx))))/(2sqrt(cos(e^((x^2)*cosx)))).`

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