Question

If a function is given as \[y=\tan (x+y)\,\], find the differentiation as \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}\]

Answer 1

Hint: Differentiate the given function thrice using sum rule and composition rule for differentiation of two functions.

Complete step-by-step solution -
We have the function \[y=\tan (x+y)\,\]. We want to find the third derivative of the given function.
We will begin by differentiating the given function with respect to \[x\]on both sides.
Thus, we have \[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \tan (x+y) \right)\,\].
We will use chain rule of composition of two functions which states that if \[y=f\left( g\left( x \right) \right)\] then we have \[\dfrac{dy}{dx}=\dfrac{df\left( g\left( x \right) \right)}{dg\left( x \right)}\times \dfrac{dg\left( x \right)}{dx}\].
Thus, we have\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \tan (x+y) \right)\,=\dfrac{d}{d(x+y)}\left( \tan (x+y) \right)\times \dfrac{d\left( x+y \right)}{dx}\]. \[...\left( 1 \right)\]
To find the value of \[\dfrac{d}{d(x+y)}\left( \tan (x+y) \right)\], let’s assume \[t=x+y\].
Thus, we have\[\dfrac{d}{d(x+y)}\left( \tan (x+y) \right)=\dfrac{d}{dt}\left( \tan t \right)\].
We know that differentiation of function of the form \[y=\tan t\] is such that\[\dfrac{d}{dt}\left( \tan t \right)={{\sec }^{2}}t\].
Thus, we have\[\dfrac{d}{d(x+y)}\left( \tan (x+y) \right)=\dfrac{d}{dt}\left( \tan t \right)={{\sec }^{2}}t={{\sec }^{2}}\left( x+y \right)\]. \[...\left( 2 \right)\]
To evaluate the value of\[\dfrac{d\left( x+y \right)}{dx}\], we will use sum rule for differentiation of two functions which states that if\[y=f\left( x \right)+g\left( x \right)\]then\[\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)\].
Thus, we have\[\dfrac{d\left( x+y \right)}{dx}=\dfrac{d}{dx}\left( x \right)+\dfrac{dy}{dx}\].
We have\[\dfrac{d}{dx}\left( x \right)=1\]as we know that the derivative of function of the form\[y=a{{x}^{n}}+b\]is\[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d\left( x+y \right)}{dx}=\dfrac{d}{dx}\left( x \right)+\dfrac{dy}{dx}=1+\dfrac{dy}{dx}\] \[...\left( 3 \right)\]
Substituting equation \[\left( 2 \right),\left( 3 \right)\]in equation\[\left( 1 \right)\], we get\[\dfrac{d}{dx}\left( \tan (x+y) \right)\,=\dfrac{d}{d(x+y)}\left( \tan (x+y) \right)\times \dfrac{d\left( x+y \right)}{dx}={{\sec }^{2}}\left( x+y \right)\left( 1+\dfrac{dy}{dx} \right)\].
Thus, we have\[\dfrac{dy}{dx}={{\sec }^{2}}(x+y)\left( 1+\dfrac{dy}{dx} \right)\].
Simplifying the above equation, we get \[\dfrac{dy}{dx}={{\sec }^{2}}(x+y)+{{\sec }^{2}}(x+y)\dfrac{dy}{dx}\]
\[\Rightarrow \dfrac{dy}{dx}\left( 1-{{\sec }^{2}}(x+y) \right)={{\sec }^{2}}(x+y)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{{{\sec }^{2}}(x+y)}{1-{{\sec }^{2}}(x+y)}=\dfrac{1+{{\tan }^{2}}(x+y)}{-{{\tan }^{2}}(x+y)}\] as we know\[{{\sec }^{2}}(x+y)=1+{{\tan }^{2}}(x+y)\]
Thus, we have \[\dfrac{dy}{dx}=\dfrac{1+{{y}^{2}}}{-{{y}^{2}}}\,=-\left( \dfrac{1}{{{y}^{2}}}+1 \right)\]as we have\[y=\tan (x+y)\,\].
\[\Rightarrow \dfrac{dy}{dx}=-\left( \dfrac{1}{{{y}^{2}}}+1 \right)\]
We will now differentiate the above equation again to find the second derivative of the given function as we know\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( -\left( 1+\dfrac{1}{{{y}^{2}}} \right) \right)\]
We can rewrite the above equation by multiplying and dividing it by\[dy\].
So, we have\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -\left( 1+\dfrac{1}{{{y}^{2}}} \right) \right)=\dfrac{d}{dy}\left( -\left( 1+\dfrac{1}{{{y}^{2}}} \right) \right)\times \dfrac{dy}{dx}\].
We know that the derivative of function of the form\[y=a{{x}^{n}}+b\]is\[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have\[\dfrac{d}{dy}\left( -\left( 1+\dfrac{1}{{{y}^{2}}} \right) \right)=\dfrac{2}{{{y}^{3}}}\].
So, we get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -\left( 1+\dfrac{1}{{{y}^{2}}} \right) \right)=\dfrac{d}{dy}\left( -\left( 1+\dfrac{1}{{{y}^{2}}} \right) \right)\times \dfrac{dy}{dx}=\dfrac{2}{{{y}^{3}}}\left( -\left( 1+\dfrac{1}{{{y}^{2}}} \right) \right)\]
Simplifying the above equation, we get\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2}{{{y}^{3}}}\left( -\dfrac{1}{{{y}^{2}}}-1 \right)=-\dfrac{2}{{{y}^{5}}}-\dfrac{2}{{{y}^{3}}}\].
To find the third derivative of the given function, we will differentiate the above equation again as\[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\].
Thus, we have\[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{d}{dx}\left( -\dfrac{2}{{{y}^{5}}}-\dfrac{2}{{{y}^{3}}} \right)\].
We will use sum rule of differentiation to get\[\dfrac{d}{dx}\left( -\dfrac{2}{{{y}^{5}}}-\dfrac{2}{{{y}^{3}}} \right)=\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{5}}} \right)+\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{3}}} \right)\]. \[...\left( 4 \right)\]
To find the value of\[\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{5}}} \right)\], rewrite the above equation by multiplying and dividing it by\[dy\].
Thus, we have\[\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{5}}} \right)=\dfrac{d}{dy}\left( \dfrac{-2}{{{y}^{5}}} \right)\times \dfrac{dy}{dx}\].
We know that\[\dfrac{d}{dy}\left( \dfrac{-2}{{{y}^{5}}} \right)=\dfrac{10}{{{y}^{6}}}\]as the derivative of function of the form\[y=a{{x}^{n}}+b\]is\[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
So, we have\[\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{5}}} \right)=\dfrac{d}{dy}\left( \dfrac{-2}{{{y}^{5}}} \right)\times \dfrac{dy}{dx}=\dfrac{10}{{{y}^{6}}}\times \dfrac{dy}{dx}\]. \[...\left( 5 \right)\]
To find the value of\[\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{3}}} \right)\], rewrite the above equation by multiplying and dividing it by\[dy\].
Thus, we have\[\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{3}}} \right)=\dfrac{d}{dy}\left( \dfrac{-2}{{{y}^{3}}} \right)\times \dfrac{dy}{dx}\].
We know that\[\dfrac{d}{dy}\left( \dfrac{-2}{{{y}^{3}}} \right)=\dfrac{6}{{{y}^{4}}}\]as the derivative of function of the form\[y=a{{x}^{n}}+b\]is\[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
So, we have\[\dfrac{d}{dx}\left( \dfrac{-2}{{{y}^{3}}} \right)=\dfrac{d}{dy}\left( \dfrac{-2}{{{y}^{3}}} \right)\times \dfrac{dy}{dx}=\dfrac{6}{{{y}^{4}}}\times \dfrac{dy}{dx}\]. \[...\left( 6 \right)\]
Substituting equation\[\left( 5 \right),\left( 6 \right)\]in equation\[\left( 4 \right)\], we get\[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{10}{{{y}^{6}}}\cdot \dfrac{dy}{dx}+\dfrac{6}{{{y}^{4}}}\dfrac{dy}{dx}\].
Substituting the value\[\dfrac{dy}{dx}=-\left( \dfrac{1}{{{y}^{2}}}+1 \right)\]in the above equation, we get\[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{10}{{{y}^{6}}}\cdot \left( -\left( \dfrac{1}{{{y}^{2}}}+1 \right) \right)+\dfrac{6}{{{y}^{4}}}\left( -\left( \dfrac{1}{{{y}^{2}}}+1 \right) \right)\]
Simplifying the above equation, we get\[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\dfrac{10}{{{y}^{8}}}-\dfrac{10}{{{y}^{6}}}-\dfrac{6}{{{y}^{6}}}-\dfrac{6}{{{y}^{4}}}=-\dfrac{10}{{{y}^{8}}}-\dfrac{16}{{{y}^{6}}}-\dfrac{6}{{{y}^{4}}}\]
Hence, we have\[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\dfrac{10}{{{y}^{8}}}-\dfrac{16}{{{y}^{6}}}-\dfrac{6}{{{y}^{4}}}\].

Note: We need to use sum rule and product rule to find the third derivative of the given functions. It’s also necessary to simplify the function as much as possible to solve the question easily.