Question

Differentiate ${{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)$ with respect to $\sqrt{1+4{{x}^{2}}}$.

Answer 1

Hint: We use the Quotient rule for Derivative of a function$f(x)$ with respect to another function $g(x)$ is given as $\dfrac{d\left( f\left( x \right) \right)}{d\left( g\left( x \right) \right)}=\dfrac{d(f(x))}{dx}\div \dfrac{d\left( g\left( x \right) \right)}{dx}$.

Complete step by step answer:
Here, we are asked to differentiate a function, say $f\left( x \right)={{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)$ with respect to $g\left( x \right)=\sqrt{1+4{{x}^{2}}}$.
i.e. we need to determine $\dfrac{d\left( f\left( x \right) \right)}{d\left( g\left( x \right) \right)}$.
We know, $\dfrac{d\left( f\left( x \right) \right)}{d\left( g\left( x \right) \right)}=\dfrac{d}{dx}f(x)\times \dfrac{dx}{d(g(x))}...........$equation$(1)$
Now, first we will find the value of the derivative of $f\left( x \right)$.
We are given $f\left( x \right)={{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)$
So, $\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}{{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)$
We can see that $f(x)$ is a composite function , i.e it is of the type $f(x)=h(p(x))$ , where $h(x)={{\tan }^{-1}}(x)$ and $p(x)=\dfrac{1+2x}{1-2x}$ . So , the derivative of $f(x)$ is given as $f'(x)=h'(p(x))\times p'(x)$
Now , we know, $\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( x \right) \right)=\dfrac{1}{1+{{x}^{2}}}$
So, $\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right) \right)=\dfrac{1}{1+{{\left( \dfrac{1+2x}{1-2x} \right)}^{2}}}.\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)........$equation$(2)$
Now, first we will determine the value of the derivative of $\dfrac{1+2x}{1-2x}$.
To find the value of the derivative of $\dfrac{1+2x}{1-2x}$, we need to apply quotient rule. The quotient rule of differentiation is given as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v{{u}^{'}}-u{{v}^{'}}}{{{v}^{2}}}$.
Now , applying quotient rule to find the derivative , we get ,
$\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)=\dfrac{\left( 1-2x \right).2-\left( -2 \right).\left( 1+2x \right)}{{{\left( 1-2x \right)}^{2}}}$
$=\dfrac{2-4x+2+4x}{{{\left( 1-2x \right)}^{2}}}$
$=\dfrac{4}{{{\left( 1-2x \right)}^{2}}}$
Now , we will substitute the value of $\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)$ in equation$(2)$.
On substituting the value of $\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)$ in equation$(2)$, we get $\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right) \right)=\dfrac{1}{1+{{\left( \dfrac{1+2x}{1-2x} \right)}^{2}}}.\dfrac{4}{{{\left( 1-2x \right)}^{2}}}$
$=\dfrac{1}{\dfrac{{{\left( 1-2x \right)}^{2}}+{{\left( 1+2x \right)}^{2}}}{{{\left( 1-2x \right)}^{2}}}}.\dfrac{4}{{{\left( 1-2x \right)}^{2}}}$
$=\dfrac{{{\left(1-2x\right)}^{2}}}{{{\left(1-2x\right)}^{2}}+{{\left(1+2x \right)}^{2}}}.\dfrac{4}{{{\left( 1-2x \right)}^{2}}}$

$=\dfrac{2}{1+4{{x}^{2}}}$
So , $\dfrac{d}{dx}f(x)=\dfrac{2}{1+4{{x}^{2}}}$
Now , we will find the derivative of$g(x)=\sqrt{1+4{{x}^{2}}}$
.
So, $\dfrac{d}{dx}g\left( x \right)=\dfrac{d}{dx}\sqrt{1+4{{x}^{2}}}$
Again $g(x)$ is a composite function. So,
$\dfrac{d}{dx}g\left( x \right)=\dfrac{1}{2\sqrt{1+4{{x}^{2}}}}.\dfrac{d}{dx}(1+4{{x}^{2}})$
$=\dfrac{4x}{\sqrt{1+4{{x}^{2}}}}$
Now, Now , we know inverse function theorem of differentiation says that if we are given $g(x)$ and $\dfrac{d}{dx}g(x)=g'(x)$ then , $\dfrac{dx}{d(g(x))}=\dfrac{1}{g'(x)}$ .
So, $\dfrac{dx}{d(g(x))}=\dfrac{1}{g'(x)}=\dfrac{1}{\dfrac{4x}{\sqrt{1+4{{x}^{2}}}}}=\dfrac{\sqrt{1+4{{x}^{2}}}}{4x}$

Now, the derivative of ${{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)$ with respect to $\sqrt{1+4{{x}^{2}}}$can be calculated by substituting the values of derivative of $\dfrac{d}{dx}f(x)$ and $\dfrac{dx}{d(g(x))}$ in equation$(1)$.
$\dfrac{d{{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)}{d\left( \sqrt{1+4{{x}^{2}}} \right)}=\dfrac{2}{1+4{{x}^{2}}}\times \dfrac{\sqrt{1+4{{x}^{2}}}}{4x}$
$=\dfrac{1}{2x\sqrt{1+4{{x}^{2}}}}$
Hence , the derivative of ${{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)$ with respect to $\sqrt{1+4{{x}^{2}}}$ is given as $\dfrac{1}{2x\sqrt{1+4{{x}^{2}}}}$
Note: While differentiating $\sqrt{1+4{{x}^{2}}}$with respect to $x$, most students make a mistake of writing $\dfrac{d}{dx}\left( \sqrt{1+4{{x}^{2}}} \right)=\dfrac{1}{2\sqrt{1+4{{x}^{2}}}}$ which is wrong.
$\dfrac{d}{dx}\sqrt{1+4{{x}^{2}}}$ $=\dfrac{1}{2\sqrt{1+4{{x}^{2}}}}\times 8x$. Such mistakes should be avoided as students can end up getting a wrong answer due to such mistakes.