Question

If$$y=\sin (x+y)$$, find $${\displaystyle \frac{d^{2}y}{d{x}^2}}$$

Answer 1

Hint: Apply chain rule or substitution method to differentiate the given trigonometric function.

Given that
$$y=\sin (x+y)$$
Differentiate the given expression w.r.t. ‘x’$$$$
$${\displaystyle \frac{dy}{dx}}=\cos (x+y)\cdot \left(1+{\displaystyle \frac{dy}{dx}}\right)$$ (Chain Rule)
$${\displaystyle \frac{dy}{dx}}=\cos (x+y)+\cos (x+y){\displaystyle \frac{dy}{dx}}$$…………. (i)
$$\Rightarrow {\displaystyle \frac{dy}{dx}}\left(1-\cos (x+y)\right)=\cos (x+y)$$
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\cos (x+y)}{1-\cos (x+y)}}$$
Differentiate the expression (i) w.r.t. ‘x’
$${\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{dy}{dx}}\right)={\displaystyle \frac{d}{dx}}\left\{\cos (x+y)+\cos (x+y)\cdot {\displaystyle \frac{dy}{dx}}\right\}$$
$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}=-\sin (x+y)\left(1+{\displaystyle \frac{dy}{dx}}\right)+\left\{-\sin (x+y)\cdot \left(1+{\displaystyle \frac{dy}{dx}}\right)\right\}{\displaystyle \frac{dy}{dx}}+\cos (x+y){\displaystyle \frac{d^{2}y}{d{x}^2}}$$
$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}\left(1-\cos (x+y)\right)=-\sin (x+y)\left(1+{\displaystyle \frac{dy}{dx}}\right)+\left\{-\sin (x+y)\cdot \left(1+{\displaystyle \frac{dy}{dx}}\right)\right\}{\displaystyle \frac{dy}{dx}}$$
$$-\sin (x+y)\left(1+{\displaystyle \frac{dy}{dx}}\right)$$ take common in R.H.S.
$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}\left(1-\cos (x+y)\right)=-\sin (x+y)\left(1+{\displaystyle \frac{dy}{dx}}\right)\left(1+{\displaystyle \frac{dy}{dx}}\right)$$
$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}\left(1-\cos (x+y)\right)=-\sin (x+y){\left(1+{\displaystyle \frac{dy}{dx}}\right)}^2$$ ……………. (ii)
Put the value of $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\cos (x+y)}{1-\cos (x+y)}}$$in expression (ii)
$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}\left(1-\cos (x+y)\right)=-\sin (x+y){\left(1+\left({\displaystyle \frac{\cos (x+y)}{1-\cos (x+y)}}\right)\right)}^2$$

$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}\left(1-\cos (x+y)\right)=-\sin (x+y){\left({\displaystyle \frac{1-\cos (x+y)+\cos (x+y)}{1-\cos (x+y)}}\right)}^2$$
$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}\left(1-\cos (x+y)\right)=-\sin (x+y){\displaystyle \frac{1}{{\left(1-\cos (x+y)\right)}^2}}$$
$$\Rightarrow {\displaystyle \frac{d^{2}y}{d{x}^2}}=-{\displaystyle \frac{\sin (x+y)}{{\left(1-\cos (x+y)\right)}^3}}$$
$$\therefore {\displaystyle \frac{d^{2}y}{d{x}^2}}=-{\displaystyle \frac{\sin (x+y)}{{\left(1-\cos (x+y)\right)}^3}}$$

Note: You can also go through with $$y=\sin (x+y)=\sin x\cdot \cos y+\cos x\cdot \sin y$$ and then differentiate.