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Question: If \[y = \sin (\sin \,x)\] , Prove that \[\dfrac{{{d^2}y}}{{d{x^2}}} + \tan \,x\,\dfrac{{dy}}{{dx}}\...

If y=sin⁡(sin⁡ x)y = \sin (\sin \,x) , Prove that d2ydx2+tan⁡ x dydx + y cos⁡2x = 0\dfrac{{{d^2}y}}{{d{x^2}}} + \tan \,x\,\dfrac{{dy}}{{dx}}\, + \,y\,{\cos ^2}x\, = \,0

Explanation

Solution

According to the given question, first apply the chain rule differentiation in the given equation y=sin⁡(sin⁡ x)y = \sin (\sin \,x). After simplifying take the double derivative of the equation and so that you can put the values in the left hand side of the equation that is d2ydx2+tan⁡ x dydx + y cos⁡2x \dfrac{{{d^2}y}}{{d{x^2}}} + \tan \,x\,\dfrac{{dy}}{{dx}}\, + \,y\,{\cos ^2}x\, . After converting all the terms in sin and cos and on simplifying we get the right hand side of the equation that is 0.

Complete step-by-step solution:
Here,  y=sin⁡(sin⁡x)      \,y = \sin \left( {\sin x} \right)\;\;\; is given in the question named as eq. (1)
To Prove:
d2ydx2+ tan⁡ x dydx+y cos⁡2x=0\dfrac{{{d^2}y}}{{d{x^2}}} + \,\tan \,x\,\dfrac{{dy}}{{dx}} + y\,{\cos ^2}x = 0
Let us differentiate the given value that is y= sin⁡(sin⁡ x)y = \,\sin (\sin \,x)
Now, on differentiating we get the value which is as follows
\Rightarrow$$$\dfrac{{dy}}{{dx}} = \cos (\sin \,x).\,\,\cos \,x$$ [Following the chain rule] On rearranging the terms we get, \Rightarrow\dfrac{{dy}}{{dx}} = \cos \,\,x.\,\,\cos (\sin \,x)$$ eq. (2) Now, double differentiating the eq. (2), we get $\Rightarrow\dfrac{{dy}}{{dx}} = \cos ,,x.,\cos (\sin ,x) Differentiating right side of the equation with respect to x, $\Rightarrow$$$\dfrac{{{d^2}y}}{{d{x^2}}} = \,( - \sin \,x).\,\cos (\sin \,x)\, + \,\cos \,x( - \sin \,(\sin \,x).\,\cos \,x)
Using the multiplication rule,
⇒\Rightarrow$$\dfrac{{{d^2}y}}{{d{x^2}}} = , - \sin ,x.,\cos (\sin ,x), - ,{\cos ^2}x.,\sin (\sin ,x)eq.(3)Now,puttingthevaluesofeq. (3) Now, putting the values of\dfrac{{dy}}{{dx}},,,\dfrac{{{d^2}y}}{{d{x^2}}}andandyfromeq.(1),eq.(2)andeq.(3)intheleft−handsideofthebelowgivenequation:Wehave,Lefthandsidetoproveas,from eq. (1), eq. (2) and eq. (3) in the left-hand side of the below given equation: We have, Left hand side to prove as, \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}, + ,\tan ,x.,,\dfrac{{dy}}{{dx}}, + ,y.,{\cos ^2}xNowwewillsubstitute, Now we will substitute,\tan ,x = \dfrac{{\sin ,x}}{{\cos ,x}} \Rightarrow , - \sin ,x.,\cos (\sin ,x), - ,,{\cos ^2}x,.\sin ,(\sin ,,x), + \dfrac{{\sin ,x}}{{\cos ,x,}},.,,\cos ,x.,,\cos (\sin ,x), + ,\sin (\sin ,x).,{\cos ^2}x,[Puttingallthevaluesfromeq.(1),eq.(2)andeq.(3)] [Putting all the values from eq. (1), eq. (2) and eq. (3)] = ,0$$
Hence, we got the value present on the R. H. S.
Hence proved.

Note: To solve these types of questions, we must remember the rules and conversion of differentiation to make the integration easier. For example differentiation of cos⁡ x\cos \,x is (−sin⁡ x)( - \sin \,x) not just sin⁡ x\sin \,x .And convert all the trigonometric values into sin and cos values that is tan⁡ x\tan \,x to sin⁡ xcos⁡ x\dfrac{{\sin \,x}}{{\cos \,x}} as this will help you to reach the correct answer.