Question

Prove that the derivative is $\dfrac{{dy}}{{dx}} = \dfrac{1}{2}$ , when $y = {\tan ^{ - 1}}\dfrac{{\sin x}}{{1 + \cos x}}$

Answer 1

Hint: We have to prove that $\dfrac{{dy}}{{dx}} = \dfrac{1}{2}$ and for that we have to find the value of $\dfrac{{dy}}{{dx}}$ and for that we will find the derivative of the function given in the question. We will differentiate the function with respect to ‘x’ and by using the appropriate formulas of trigonometric functions.

Complete step-by-step solution:
Given: $y = {\tan ^{ - 1}}\dfrac{{\sin x}}{{1 + \cos x}}$
By using the formula $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $\cos x = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}$ we will rewrite the function given above.
\[y = {\tan ^{ - 1}}\dfrac{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{1 + {{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}\]
By using the trigonometric formula we can write ${\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}$ as $2{\cos ^2}\dfrac{x}{2} - 1$ .
So, replacing it in the denominator.
\[y = {\tan ^{ - 1}}\dfrac{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{1 + 2{{\cos }^2}\dfrac{x}{2} - 1}}\]
\[y = {\tan ^{ - 1}}\dfrac{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}\]
Now, we will take $2\cos \dfrac{x}{2}$ common from numerator and denominator and cut them from each other.
\[y = {\tan ^{ - 1}}\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}\]
By using the trigonometric formula $\dfrac{{\sin A}}{{\cos A}} = \tan A$ we will rewrite the function.
\[y = {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)\]
Now, according to trigonometric properties\[{\tan ^{ - 1}}\left( {\tan A} \right)\] is equal to $A$ . So, replacing it in the equation above.
$y = \dfrac{x}{2}$
We will now differentiate the equation above with respect to ‘x’ to find the value of $\dfrac{{dy}}{{dx}}$ .
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}$
We know that the derivative of $x$ is $1$. So, the derivative of $\dfrac{x}{2}$ will be $\dfrac{1}{2}$ .
Hence, proved.

Note: Differentiation is a method of finding the derivative of function. Differentiation is a process, in Maths, where we find the instantaneous rate of change in function based on one of its variables. The most common example is the rate change of displacement with respect to time, called velocity. The opposite of finding a derivative is anti-differentiation.