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Question: Differentiate \[{\sin ^2}\left( {{x^2}} \right)\] with respect to \[{x^2}\]...

Differentiate sin2(x2){\sin ^2}\left( {{x^2}} \right) with respect to x2{x^2}

Explanation

Solution

:We will make use of the standard formula which says As we know that 2 sinx . cosx = sin 2x2{\text{ }}sinx{\text{ }}.{\text{ }}cosx{\text{ }} = {\text{ }}sin{\text{ }}2x. Further we will differentiate the value sin2(x2){\sin ^2}({x^2}).

Complete step by step solution:
Consider the given differentiation
y=sin2(x2)y = {\sin ^2}\left( {{x^2}} \right)
Differentiate with respect to x2{x^2}
dydx2=dsin2x2dx2\dfrac{{dy}}{{d{x^2}}} = \dfrac{{d{{\sin }^2}{x^2}}}{{d{x^2}}}
dydx2=2sinx2cosx2\dfrac{{dy}}{{d{x^2}}} = 2\sin {x^2}\cos {x^2}
As we know that 2 sin xcos x=sin 2x so,2{\text{ }}sin{\text{ }}xcos{\text{ }}x = sin{\text{ }}2x{\text{ }}so,
2sinx2cosx2=sin2x22\sin {x^2}\cos {x^2} = \sin 2{x^2}
dydx2=sin(2x2)\dfrac{{dy}}{{d{x^2}}} = \sin \left( {2{x^2}} \right)
Hence, this is the answer

Note: We can also solve these problem by taking f=sin2x2g=x2f = {\sin ^2}{x^2}\,\,g = {x^2} to get
dfdg=dfdb×dbdg=df/dbdb/dg\dfrac{{df}}{{dg}} = \dfrac{{df}}{{db}} \times \dfrac{{db}}{{dg}} = \dfrac{{df/db}}{{db/dg}}