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Question: If \(y = \sin \left( {{{\log }_e}x} \right)\), then prove that \(\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt ...

If y=sin(logex)y = \sin \left( {{{\log }_e}x} \right), then prove that dydx=1y2x\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {y^2}} }}{x}.

Explanation

Solution

Here, will differentiate both the sides of the given equation with respect to xx. Then, we will use different trigonometric identities to rewrite the trigonometric value present in the differentiation such that we will be able to prove the given equation.

Formula Used:
cosθ=1sin2θ\cos \theta = \sqrt {1 - {{\sin }^2}\theta }

Complete step-by-step answer:
According to the question,
y=sin(logex)y = \sin \left( {{{\log }_e}x} \right)……………………..(1)\left( 1 \right)
Now, differentiating both sides with respect to xx using the formula dydxsinx=cosx\dfrac{{dy}}{{dx}}\sin x = \cos x and dydxlogx=1x\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}, we get
dydx=cos(logex)×dydx(logex)\Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {{{\log }_e}x} \right) \times \dfrac{{dy}}{{dx}}\left( {{{\log }_e}x} \right)
dydx=cos(logex)×1x\Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {{{\log }_e}x} \right) \times \dfrac{1}{x}…………………………….(2)\left( 2 \right)
Now, we know that, cos2θ+sin2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
This can also be written as:
cosθ=1sin2θ\Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta }
Now, substituting θ=logex\theta = {\log _e}x in cosθ=1sin2θ\cos \theta = \sqrt {1 - {{\sin }^2}\theta } , we get
cos(logex)=1sin2(logex)\cos \left( {{{\log }_e}x} \right) = \sqrt {1 - {{\sin }^2}\left( {{{\log }_e}x} \right)}
Substituting this in (2)\left( 2 \right), we get,
dydx=1sin2(logex)×1x\dfrac{{dy}}{{dx}} = \sqrt {1 - {{\sin }^2}\left( {{{\log }_e}x} \right)} \times \dfrac{1}{x}………………………………….(3)\left( 3 \right)
From equation (1)\left( 1 \right), we know that y=sin(logex)y = \sin \left( {{{\log }_e}x} \right)
Squaring both sides,
y2=sin2(logex)\Rightarrow {y^2} = {\sin ^2}\left( {{{\log }_e}x} \right)
Now substituting this in equation (3)\left( 3 \right), we get
dydx=1y2x\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {y^2}} }}{x}
Hence, proved

Note: In this question, we have used trigonometry identity to prove this question. Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by the cartographers to make maps. In fact, trigonometry is even used by Astronomers to find the distance between two stars. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In the simple terms they are written as ‘sin’, ‘cos’ and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.