Question

If $x\cos (a+y)=\cos y$, then prove that $\dfrac{dy}{dx}=\dfrac{{{\cos }^{2}}(a+y)}{\sin a}$. Show that $\sin a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\sin 2(a+y)\dfrac{dy}{dx}=0$.

Answer 1

Before solving this question we should be aware of derivatives of trigonometric functions. For solving this question we will derive the given equation with respect to $x$,2 times and simplify the equations.

Complete step by step answer:
$\begin{aligned} & \dfrac{d}{dx}\cos x=-\sin x \\\ & \sin (A-B)=\sin A\cos B-\sin B\cos A \\\ \end{aligned}$
We are going to use the above formulas and chain rule method for solving this question.
Chain rule method: $\dfrac{d}{dx}u=\dfrac{du}{dy}.\dfrac{dy}{dx}$ where $u$ is a function of $y$.
Let’s derivative the given equation with respect to $x$
We have $x=\dfrac{\cos y}{\cos (a+y)}$
By applying derivation with respect to $x$ on both sides.
$\begin{aligned} & \dfrac{dx}{dx}=\dfrac{d}{dy}\left( \dfrac{\cos y}{\cos (a+y)} \right)\dfrac{dy}{dx} \\\ & \\\ \end{aligned}$
By using chain rule, we have got the above simplified equation.
As we know$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ , $\dfrac{d}{dx}\cos x=-\sin x$and $\sin (A-B)=\sin A\cos B-\sin B\cos A$
Here, we have $\dfrac{d}{dy}\cos (a+y)$ to simplify this let us assume $u=a+y$
And perform some simplifications.
$\begin{aligned} & \dfrac{du}{dy}=0+\dfrac{dy}{dy} \\\ & du=dy \\\ \end{aligned}$
Let us replace $du$ with $dy$.
Since $a$ is constant.
$\begin{aligned} & \dfrac{d}{du}\cos u=-\sin u \\\ & \Rightarrow \dfrac{d}{dx}\cos (a+y)=-\sin (a+y) \\\ \end{aligned}$
Using this in the above equation, we get
$\begin{aligned} & 1=\dfrac{\cos (a+y)(-\sin y)-\cos y(-\sin (a+y))}{{{\cos }^{2}}(a+y)}\cdot \dfrac{dy}{dx} \\\ & 1=\dfrac{\sin (a+y-y)}{{{\cos }^{2}}(a+y)}\cdot \dfrac{dy}{dx} \\\ & \dfrac{dy}{dx}=\dfrac{{{\cos }^{2}}(a+y)}{\sin a} \\\ \end{aligned}$
Hence, proved.
Now, we required to prove that $\sin a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\sin 2(a+y)\dfrac{dy}{dx}=0$.
Let's derive the value we got once again with respect to $x$.
$\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{{{\cos }^{2}}(a+y)}{\sin a} \right) \\\ & \sin a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( {{\cos }^{2}}(a+y) \right) \\\ \end{aligned}$
Since $a$ is constant $\sin a$ is also constant and $\dfrac{d}{dx}\sin a=0$.
As we know $\dfrac{d}{dx}{{\cos }^{2}}x=-\sin 2x$.
Here, we have $\dfrac{d}{dx}{{\cos }^{2}}(a+y)$ to simplify this let us assume $u=a+y$
And perform some simplifications.
$\begin{aligned} & \dfrac{du}{dy}=0+\dfrac{dy}{dy} \\\ & du=dy \\\ \end{aligned}$
Let us replace $du$ with $dy$.
Since $a$ is constant.
$\begin{aligned} & \dfrac{d}{du}{{\cos }^{2}}u=-\sin 2u \\\ & \Rightarrow \dfrac{d}{dy}{{\cos }^{2}}(a+y)=-\sin 2(a+y) \end{aligned}$
Using this in the above equation, let us simplify the equation we have
$\begin{aligned} & \sin a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left[ \dfrac{d}{dy}{{\cos }^{2}}(a+y) \right]\dfrac{dy}{dx} \\\ & \sin a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=[-\sin 2(a+y)]\dfrac{dy}{dx} \\\ & \\\ \end{aligned}$
$\sin a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\sin 2(a+y)\dfrac{dy}{dx}=0$
Hence, proved.

Note: Here in this question,$a$ is constant. So,$\sin a$ is also constant and $\dfrac{d}{dx}\sin a=0$.If we take $a$ as not constant it will lead us to complete different conclusion and we will also face difficulty will trying to simplify the equations we have. Because of mistakes like these we will be directed towards a completely different answer.