Question

Differentiate the following function with respect to\[x\]:$\sqrt{\tan \sqrt{x}}$.

Answer 1

Hint: Let$y=\sqrt{\tan \sqrt{x}}$, then differentiate y with respect to $x$.
First of all let us see what is the differentiation of $\tan x$and $\sqrt{x}$.
Then we have
$\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$
$\dfrac{d\sqrt{x}}{dx}=\dfrac{d{{\left( x \right)}^{\dfrac{1}{2}}}}{dx}$
As we know ${{\dfrac{d\left( x \right)}{dx}}^{n}}=n{{x}^{n-1}}$
Therefore we have, $\dfrac{d{{\left( x \right)}^{\dfrac{1}{2}}}}{dx}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}$ $=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$.

Complete step by step answer:
Now we will find the differentiation of $\sqrt{\tan \sqrt{x}}$with respect to$x$.
For this, let us consider
$y=\sqrt{\tan \sqrt{x}}$
Differentiating y with respect to$x$, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\sqrt{\tan \sqrt{x}}$
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\tan \sqrt{x}}.\dfrac{d\left( \tan \sqrt{x} \right)}{dx}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\tan \sqrt{x}}.{{\sec }^{2}}\sqrt{x}.\dfrac{d\left( \sqrt{x} \right)}{dx}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\tan \sqrt{x}}.{{\sec }^{2}}\sqrt{x}.\dfrac{1}{2\sqrt{x}}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{{{\sec }^{2}}\sqrt{x}}{4\sqrt{x}\tan \sqrt{x}}\]

$\therefore \dfrac{d\left( \sqrt{\tan \sqrt{x}} \right)}{dx}=\dfrac{{{\sec }^{2}}\sqrt{x}}{4\sqrt{x}\tan \sqrt{x}}$

Hence, differentiation of $\sqrt{\tan \sqrt{x}}$with respect to $x$ is $\dfrac{d\left( \sqrt{\tan \sqrt{x}} \right)}{dx}=\dfrac{{{\sec }^{2}}\sqrt{x}}{4\sqrt{x}\tan \sqrt{x}}$.

Note: The basic differentiation rules that need to be followed are:
(a) Sum or difference rule – If the function is sum or difference of two function, then the derivatives of the function is the sum or difference of the individual functions, i.e., if $x=y\pm z$, then $\dfrac{dx}{dt}=\dfrac{dy}{dt}\pm \dfrac{dz}{dt}$.
(b) Product rule – As per the product rule, if function $x$is the product of two functions $y$ and$z$, then the derivative of the function is as below.
If$x=yz$, then
\[\dfrac{dx}{dt}=\dfrac{dy}{dt}.z+\dfrac{dz}{dt}.y\]
(c) Quotient rule – If the function $x$ is in the form two functions$\dfrac{y}{z}$, then the derivative of the function is as below.
If $x=\dfrac{y}{z}$, then
$\dfrac{dx}{dt}=\dfrac{\dfrac{dy}{dt}.z-\dfrac{dz}{dt}.y}{{{z}^{2}}}$ .
(d) Chain rule – If a function $y=f\left( x \right)=g\left( u \right)$ and if$u=h\left( x \right)$, then the chain rule for differentiation is defined as,
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$.
This plays a major role in the method of substitution that helps to perform differentiation of composite functions.