InterviewSolution
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InterviewSolution
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`y=e^(sin x^(3))` |
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Answer» Correct Answer - `e^(sin x^(2))cos x^(2)2x` `"Let "y=e^(sinx^(2)).` Putting `x^(2)=v and u=sin x^(2) = sin v,` we get `y=e^(u),u=sin v, and v=x^(2)` `therefore" "(dy)/(du)=e^(u),(du)/(dv)=cos v, and (dv)/(dx)=2x` `"Now, "(dy)/(dx)xx(dy)/(du)xx(du)/(dv)xx(dv)/(dx)` `=e^(u)cos v 2=e^(sin v)cos v 2x` `=e^(sin x^(2))cos x^(2)2x` |
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