Question

Find the derivative of cos inverse (x-1/x)/(x+1/x) by proper substitution method

Answer 1

Let,

$\quad$

$\displaystyle\longrightarrow\sf {f(x)=\cos^{-1}\left (\dfrac {x-\frac {1}{x}}{x+\frac {1}{x}}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {f(x)=\cos^{-1}\left (\dfrac {\frac {x^2-1}{x}}{\frac {x^2+1}{x}}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {f(x)=\cos^{-1}\left (\dfrac {x^2-1}{x^2+1}\right)\quad\quad\dots (1)}$

$\quad$

Let,

$\quad$

$\displaystyle\mapsto\sf {g(x)=\dfrac {x^2-1}{x^2+1}}$

$\quad$

Then (1) becomes,

$\quad$

$\displaystyle\longrightarrow\sf {f(x)=\cos^{-1}\left (g(x)\right)}$

$\quad$

Taking the derivative wrt $\displaystyle\sf {x,}$

$\quad$

$\displaystyle\longrightarrow\sf {f'(x)=\dfrac {d[\cos^{-1}\left (g(x)\right)]}{dx}}$

$\quad$

By chain rule,

$\quad$

$\displaystyle\longrightarrow\sf {f'(x)=\dfrac {d[\cos^{-1}\left (g(x)\right)]}{d[g(x)]}\cdot\dfrac {d[g(x)]}{dx}\quad\quad\dots (2)}$

$\quad$

Well,

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[\cos^{-1}\left (g(x)\right)]}{d[g(x)]}=-\dfrac {1}{\sqrt{1-(g(x))^2}}}$

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[\cos^{-1}\left (g(x)\right)]}{d[g(x)]}=-\dfrac {1}{\sqrt{1-\left(\dfrac {x^2-1}{x^2+1}\right)^2}}}$

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[\cos^{-1}\left (g(x)\right)]}{d[g(x)]}=-\dfrac {1}{\sqrt{\dfrac {(x^2+1)^2-(x^2-1)^2}{(x^2+1)^2}}}}$

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[\cos^{-1}\left (g(x)\right)]}{d[g(x)]}=-\dfrac {x^2+1}{\sqrt{(x^2+1)^2-(x^2-1)^2}}}$

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[\cos^{-1}\left (g(x)\right)]}{d[g(x)]}=-\dfrac {x^2+1}{2x}}$

$\quad$

And,

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[g(x)]}{dx}=\dfrac {d}{dx}\left (\dfrac {x^2-1}{x^2+1}\right)}$

$\quad$

By quotient rule,

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[g(x)]}{dx}=\dfrac {2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}}$

$\quad$

$\displaystyle\mapsto\sf {\dfrac {d[g(x)]}{dx}=\dfrac {4x}{(x^2+1)^2}}$

$\quad$

Then (2) becomes,

$\quad$

$\displaystyle\longrightarrow\sf {f'(x)=-\dfrac {x^2+1}{2x}\cdot\dfrac {4x}{(x^2+1)^2}}$

$\quad$

$\displaystyle\longrightarrow\sf {f'(x)=-\dfrac {2}{x^2+1}}$

$\quad$

That is,

$\quad$

$\displaystyle\longrightarrow\sf {\underline {\underline {\dfrac {d}{dx}\left [\cos^{-1}\left (\dfrac {x-\frac {1}{x}}{x+\frac {1}{x}}\right)\right]=-\dfrac {2}{x^2+1}}}}$