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

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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’

\[\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left\{ {\cos (x + y) + \cos (x + y)

\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.

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’

\[\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left\{ {\cos (x + y) + \cos (x + y)

\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.

Last updated date: 26th Sep 2023

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