Question

Given if $\sin y=x\sin \left( a+y \right)$, prove the result $\dfrac{dy}{dx}=\dfrac{{{\sin }^{2}}\left( a+y \right)}{\sin a}$.

Answer 1

We start solving the problem by finding the value of from the given equation of the problem $\sin y=x\sin \left( a+y \right)$. We use the $\dfrac{u}{v}$ rule of differentiation to differentiate x with respect to y. We use the results of differentiation and the result of $\sin \left( A-B \right)$, to get the value of $\dfrac{dx}{dy}$. Now, we take the inverse to get the required result.

Complete step by step answer:
Given that we have $\sin y=x\sin \left( a+y \right)$ and we need to prove the result $\dfrac{dy}{dx}=\dfrac{{{\sin }^{2}}\left( a+y \right)}{\sin a}$.
We have got the value $\sin y=x\sin \left( a+y \right)$.
We have got the value $x=\dfrac{\sin y}{\sin \left( a+y \right)}$.
Let us differentiate this equation on both sides.
We have got the value $\dfrac{dx}{dy}=\dfrac{d}{dy}\left( \dfrac{\sin y}{\sin \left( a+y \right)} \right)$ ---(1).
We know that differentiation of a function $\dfrac{u}{v}$ is defined as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$. We use this result in equation (1).
We have got the value $\dfrac{dx}{dy}=\left( \dfrac{\left( \sin \left( a+y \right)\times \dfrac{d}{dy}\left( \sin y \right) \right)-\left( \sin y\times \dfrac{d}{dy}\left( \sin \left( a+y \right) \right) \right)}{{{\sin }^{2}}\left( a+y \right)} \right)$ ---(2).
We know that $\dfrac{d}{dx}\left( \sin x \right)=\cos x$ and $\dfrac{d}{dx}\left( \sin \left( f\left( x \right) \right) \right)=\cos \left( f\left( x \right) \right)\times \dfrac{d}{dx}\left( f\left( x \right) \right)$. We use this result in equation (2).
We have got the value $\dfrac{dx}{dy}=\left( \dfrac{\left( \sin \left( a+y \right)\times \cos y \right)-\left( \sin y\times \left( \cos \left( a+y \right)\times \dfrac{d\left( a+y \right)}{dy} \right) \right)}{{{\sin }^{2}}\left( a+y \right)} \right)$.
We have got the value $\dfrac{dx}{dy}=\left( \dfrac{\left( \sin \left( a+y \right)\times \cos y \right)-\left( \sin y\times \left( \cos \left( a+y \right)\times 1 \right) \right)}{{{\sin }^{2}}\left( a+y \right)} \right)$.
We have got the value $\dfrac{dx}{dy}=\left( \dfrac{\left( \sin \left( a+y \right)\times \cos y \right)-\left( \sin y\times \cos \left( a+y \right) \right)}{{{\sin }^{2}}\left( a+y \right)} \right)$ ---(3).
We know that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$. We use this result in equation (3).
We have got the value $\dfrac{dx}{dy}=\left( \dfrac{\sin \left( a+y-y \right)}{{{\sin }^{2}}\left( a+y \right)} \right)$.
We have got the value $\dfrac{dx}{dy}=\left( \dfrac{\sin \left( a \right)}{{{\sin }^{2}}\left( a+y \right)} \right)$ ---(4).
We know that if $\dfrac{dy}{dx}=a$, then $\dfrac{dx}{dy}=\dfrac{1}{a}$. We use this result in equation (4).
We have got the value $\dfrac{dy}{dx}=\dfrac{{{\sin }^{2}}\left( a+y \right)}{\sin \left( a \right)}$.
We have found the value of $\dfrac{dy}{dx}$ as $\dfrac{{{\sin }^{2}}\left( a+y \right)}{\sin \left( a \right)}$.

∴ We have proved the result $\dfrac{dy}{dx}=\dfrac{{{\sin }^{2}}\left( a+y \right)}{\sin \left( a \right)}$.

Note: We should not make mistakes while differentiating using $\dfrac{u}{v}$ rule. We should not forget that we can use inverse after differentiation also. Because sometimes finding $\dfrac{dy}{dx}$ directly will be tough when compared to finding $\dfrac{dx}{dy}$.