Question

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

Answer 1

Hint: According to the given question, first apply the chain rule differentiation in the given equation \[y = \sin (\sin \,x)\]. 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 \[\dfrac{{{d^2}y}}{{d{x^2}}} + \tan \,x\,\dfrac{{dy}}{{dx}}\, + \,y\,{\cos ^2}x\,\] . 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, \[\,y = \sin \left( {\sin x} \right)\;\;\;\] is given in the question named as eq. (1)
To Prove:
\[\dfrac{{{d^2}y}}{{d{x^2}}} + \,\tan \,x\,\dfrac{{dy}}{{dx}} + y\,{\cos ^2}x = 0\]
Let us differentiate the given value that is \[y = \,\sin (\sin \,x)\]
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, putting the values of \[\dfrac{{dy}}{{dx}}\,,\,\dfrac{{{d^2}y}}{{d{x^2}}}\] and \[y\] 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,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\, + \,\tan \,x.\,\,\dfrac{{dy}}{{dx}}\, + \,y.\,{\cos ^2}x\]
Now we will substitute, \[\tan \,x = \dfrac{{\sin \,x}}{{\cos \,x}}\]
\[ \Rightarrow \, - \sin \,x.\,\cos (\sin \,x)\, - \,\,{\cos ^2}x\,.\sin \,(\sin \,\,x)\, + \dfrac{{\sin \,x}}{{\cos \,x\,}}\,.\,\,\cos \,x.\,\,\cos (\sin \,x)\, + \,\sin (\sin \,x).\,{\cos ^2}x\,\]
 [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 \[\cos \,x\] is \[( - \sin \,x)\] not just \[\sin \,x\] .And convert all the trigonometric values into sin and cos values that is \[\tan \,x\] to \[\dfrac{{\sin \,x}}{{\cos \,x}}\] as this will help you to reach the correct answer.