Question

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

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’$$$$
$\dfrac{{dy}}{{dx}} = \cos (x + y) \cdot \left( {1 + \dfrac{{dy}}{{dx}}} \right)$ (Chain Rule)
$\dfrac{{dy}}{{dx}} = \cos (x + y) + \cos (x + y)\dfrac{{dy}}{{dx}}$…………. (i)
$\Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - \cos (x + y)} \right) = \cos (x + y)$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos (x + y)}}{{1 - \cos (x + y)}}$
Differentiate the expression (i) w.r.t. ‘x’

\cdot \dfrac{{dy}}{{dx}}} \right\\}$$ $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \sin (x + y)\left( {1 + \dfrac{{dy}}{{dx}}} \right) + \left\\{ { - \sin (x + y) \cdot \left( {1 + \dfrac{{dy}}{{dx}}} \right)} \right\\}\dfrac{{dy}}{{dx}} + \cos (x \+ y)\dfrac{{{d^2}y}}{{d{x^2}}}$$ $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\left( {1 - \cos (x + y)} \right) = - \sin (x + y)\left( {1 + \dfrac{{dy}}{{dx}}} \right) + \left\\{ { - \sin (x + y) \cdot \left( {1 + \dfrac{{dy}}{{dx}}} \right)} \right\\}\dfrac{{dy}}{{dx}}$$ $$ - \sin (x + y)\left( {1 + \dfrac{{dy}}{{dx}}} \right)$$ take common in R.H.S. $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\left( {1 - \cos (x + y)} \right) = - \sin (x + y)\left( {1 + \dfrac{{dy}}{{dx}}} \right)\left( {1 + \dfrac{{dy}}{{dx}}} \right)$$ $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\left( {1 - \cos (x + y)} \right) = - \sin (x + y){\left( {1 + \dfrac{{dy}}{{dx}}} \right)^2}$$ ……………. (ii) Put the value of $$\dfrac{{dy}}{{dx}} = \dfrac{{\cos (x + y)}}{{1 - \cos (x + y)}}$$in expression (ii) $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\left( {1 - \cos (x + y)} \right) = - \sin (x + y){\left( {1 + \left( {\dfrac{{\cos (x + y)}}{{1 - \cos (x + y)}}} \right)} \right)^2}$$ $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\left( {1 - \cos (x + y)} \right) = - \sin (x + y){\left( {\dfrac{{1 - \cos (x + y) + \cos (x + y)}}{{1 - \cos (x + y)}}} \right)^2}$$ $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\left( {1 - \cos (x + y)} \right) = - \sin (x + y)\dfrac{1}{{{{\left( {1 - \cos (x + y)} \right)}^2}}}$$ $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{\sin (x + y)}}{{{{\left( {1 - \cos (x + y)} \right)}^3}}}$$ $$\therefore \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{\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.