Question

What is the derivative of $\arctan \left( {\dfrac{{\sqrt x - x}}{{1 + {x^{1.5}}}}} \right)$

Answer 1

Hint : $\arctan (x)$ is also written as ${\tan ^{ - 1}}x$ . You should first simplify this question using the expansion formula of inverse functions. And then differentiate using formulae of derivatives.

Complete step-by-step answer :
$\arctan \left( {\dfrac{{\sqrt x - x}}{{1 + {x^{1.5}}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt x - x}}{{1 + {x^{1.5}}}}} \right)$
It can further be written as
$= {\tan ^{ - 1}}\left( {\dfrac{{\sqrt x - x}}{{1 + {x^{\dfrac{3}{2}}}}}} \right)$
We can write it in terms of ${\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$ by writing it as
$= {\tan ^{ - 1}}\left( {\dfrac{{\sqrt x - x}}{{1 + {x^{\dfrac{1}{2}}}(x)}}} \right)$ $\left( {\because {a^n} \times {a^m} = {a^{nm}}} \right)$
$= {\tan ^{ - 1}}\left( {\dfrac{{\sqrt x - x}}{{1 + \sqrt x x}}} \right)$ $\left( {\because {a^{\dfrac{1}{2}}} = \sqrt a } \right)$
Now it is of the form ${\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$
So we can use the formula ${\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y$ to simplify it
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\sqrt x - x}}{{1 + {x^{1.5}}}}} \right) = {\tan ^{ - 1}}\sqrt x - {\tan ^{ - 1}}x$
Now since we have simplified a question. Now we should differentiate it with respect to $x$ to find the derivative of the above equation.
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{{\sqrt x - x}}{{1 + {x^{1.5}}}}} \right)} \right) = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\sqrt x - {{\tan }^{ - 1}}x} \right)$
Since, derivative is distributive over addition and subtraction, we can write the above expression as
$\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\sqrt x - {{\tan }^{ - 1}}x} \right) = \dfrac{d}{{dx}}{\tan ^{ - 1}}\sqrt x - \dfrac{d}{{dx}}{\tan ^{ - 1}}x$
We know that
$\dfrac{d}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$ and $\dfrac{d}{{dx}}f(g(x)) = f'(g(x))\dfrac{d}{{dx}}(g(x))$
Using these identities, we get
$\dfrac{d}{{dx}}{\tan ^{ - 1}}\sqrt x - \dfrac{d}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + x}}\dfrac{d}{{dx}}\sqrt x - \dfrac{1}{{1 + {x^2}}}$
$= \dfrac{1}{{1 + x}} \times \dfrac{1}{{2\sqrt x }} - \dfrac{1}{{1 + {x^2}}}$ $\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$
$= \dfrac{1}{{(1 + x)2\sqrt x }} - \dfrac{1}{{1 + {x^2}}}$
Therefore, derivative of $\arctan \left( {\dfrac{{\sqrt x - x}}{{1 + {x^{1.5}}}}} \right)$ is $\dfrac{1}{{(1 + x)2\sqrt x }} - \dfrac{1}{{1 + {x^2}}}$

Note : You should never try to solve such questions by differentiating them directly. Direct differentiation will lead to complex calculations where you will be prone to making mistakes. So it is better to simplify the question first to a form that is comparatively easy to differentiate. Like we did in this question by simplifying $\arctan \left( {\dfrac{{\sqrt x - x}}{{1 + {x^{1.5}}}}} \right)$ to ${\tan ^{ - 1}}\sqrt x - {\tan ^{ - 1}}x$ and then we differentiated it. You need to know the expansion formulae and you need to read the question carefully to understand if it can be converted into the form of the expansion formula. Like in this question ${x^{1.5}}$ was written instead of $x\sqrt x$ . You should be able to pinpoint to that.