Question

If $ \sqrt {\tan y} = {e^{\cos 2x}}\sin x $ , then find $ \dfrac{{dy}}{{dx}} $ .

Answer 1

Hint: In the given question, we are given an equation involving two variables and we have to differentiate both sides of the equation with respect to x, in order to find the differential $ \dfrac{{dy}}{{dx}} $ . The question involves the concepts of differentiation and its rules. Hence, we must remember rules of differentiation in order to solve such questions.

Complete step by step solution:
 $ \sqrt {\tan y} = {e^{\cos 2x}}\sin x $
Squaring both sides of the equation, we get,
 $ \Rightarrow \tan y = {\left( {{e^{\cos 2x}}\sin x} \right)^2} $
 $ \Rightarrow \tan y = {e^{2\cos 2x}}{\sin ^2}x - - - - - \left( 1 \right) $
Now, differentiating both sides of the equation with respect to x, we get,
We know that $ {\sec ^2}x $ is the derivative of $ \tan x $ . Also, using product rule, $ \dfrac{d}{{dx}}\left[ {f\left( x \right).g\left( x \right)} \right] = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right) $ , we get,
 $ \Rightarrow {\sec ^2}y\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left[ {{e^{2\cos 2x}}{{\sin }^2}x} \right] $
Also, using chain rule of differentiation, $ \dfrac{d}{{dx}}\left[ {f\left[ {g\left( x \right)} \right]} \right] = f'\left[ {g\left( x \right)} \right] \times g'\left( x \right) $ , we get,
 $ \Rightarrow {\sec ^2}y\left( {\dfrac{{dy}}{{dx}}} \right) = {\sin ^2}x\left( {{e^{2\cos 2x}}} \right)\left( 2 \right)\left( { - \sin 2x} \right)\left( 2 \right) + {e^{2\cos 2x}}\left( {2\sin x\cos x} \right) $
Simplifying the expression further, we get,
 $ \Rightarrow {\sec ^2}y\left( {\dfrac{{dy}}{{dx}}} \right) = - 4{e^{2\cos 2x}}\sin 2x{\sin ^2}x + {e^{2\cos 2x}}\sin 2x $
Taking $ {e^{2\cos 2x}}\sin 2x $ common in right side of the equation, we get,
 $ \Rightarrow {\sec ^2}y\left( {\dfrac{{dy}}{{dx}}} \right) = {e^{2\cos 2x}}\sin 2x\left[ {1 - 4{{\sin }^2}x} \right] $
Taking all the parameters to right side of the equation except the differential $ \dfrac{{dy}}{{dx}} $ ,
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = {e^{2\cos 2x}}{\cos ^2}y\sin 2x\left[ {1 - 4{{\sin }^2}x} \right]\]
Now, from equation $ \left( 1 \right) $ , we have, $ \tan y = {e^{2\cos 2x}}{\sin ^2}x $ ,
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( {{e^{2\cos 2x}}{{\sin }^2}x} \right){\cos ^2}y\left( {\dfrac{{2\cos x}}{{\sin x}}} \right)\left[ {1 - 4{{\sin }^2}x} \right]\]
Rearranging the terms a bit to simplify the expression and match the options, we get,
Substituting $ \tan y $ for $ {e^{2\cos 2x}}{\sin ^2}x $ , we have,
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \tan y{\cos ^2}y\left( {\dfrac{{2\cos x}}{{\sin x}}} \right)\left[ {1 - 4{{\sin }^2}x} \right]\]
Now, simplifying the expression further by writing $ \tan y $ as $ \dfrac{{\sin y}}{{\cos y}} $ , we get,
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{2\sin y}}{{\cos y}}{\cos ^2}y\left( {\dfrac{{\cos x}}{{\sin x}}} \right)\left[ {1 - 4{{\sin }^2}x} \right]\]
Condensing the expression using double angle formula of sine, \[2\sin y\cos y = \sin 2y\] , we get,
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = 2\sin y\cos y\left( {\dfrac{{\cos x}}{{\sin x}}} \right)\left[ {1 - 4{{\sin }^2}x} \right]\]
Multiplying $ \cot x $ within the bracket, we get,
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \sin 2y\left[ {\cot x - 4\cos x\sin x} \right]\]
Again using the double angle formula for sine, we get,
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \sin 2y\left[ {\cot x - 2\sin 2x} \right]\]
Hence, option (a) is correct.
Therefore, if $ \sqrt {\tan y} = {e^{\cos 2x}}\sin x $ , then $ \dfrac{{dy}}{{dx}} = \sin 2y\left[ {\cot x - 2\sin 2x} \right] $ .
So, the correct answer is “$ \sin 2y\left[ {\cot x - 2\sin 2x} \right] $ . ”.

Note: Such questions require prerequisite knowledge of differentiation. Also, basic knowledge of algebra and simplification rules is also of vital importance in handling complex and tedious steps in such a problem.