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How do you find the derivative of $f(x)=\sqrt{\sin x\left( 2x \right)}?$

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Last updated date: 13th Jun 2024
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Answer
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Hint: As derivative of a function of real variable measures the sensitivity to change of the function value with respect to change in its argument. Derivatives are a fundamental tool of calculus.
Use chain rule to find the derivative of $f(x)=\sqrt{\sin \left( 2x \right)}$
Chain Rule:- $fg'(x).g'(x)$
Example: $\sin (5x)$ a competitive function.
$f(x)=\sin x\Rightarrow f'(x)=\cos x$
$g(x)=\sin x\Rightarrow g'(x)=5$
So, the derivative will be equal to
$\cos \left( 5x \right)5$

Complete step by step solution:
You know that, given function is
$f(x)=\sqrt{\sin \left( 2x \right)}$
Firstly, let $y=\sqrt{\sin \left( 2x \right)}$
And let $u=\sin \left( 2x \right)$
This mean $y={{u}^{\dfrac{1}{2}}}$
Therefore,
$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}{{u}^{\dfrac{-1}{2}}}.2\cos \left( 2x \right)$
Which implies, as ${{u}^{\dfrac{-1}{2}}}=-\sqrt{u}$ therefore as you transfer to denominator it gets reprobated into $\sqrt{4}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}.\dfrac{2\cos 2x}{\sqrt{u}}$
And $'2'$ of denominator and $2$ of $'2\cos 2x'$ gets canceled.
Therefore, simplified equation will be
$\dfrac{dy}{dx}=\dfrac{\cos 2x}{\sqrt{u}}$
Replace $u=\sin 2x$ to get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\cos 2x}{\sqrt{\sin 2x}}$

Additional Information:
The chain rule tells us how to find the derivative of composite function.
The chain rule, formula is,
$\dfrac{d}{dx}\left[ f\left( f(x) \right) \right]=f'(g(x)g'(x))$
A function is composite if you can write it as $f(g(x))$. In other words, it is a function within a function, or a function of function. For example: $\cos \left( {{x}^{2}} \right)$ is composite because if we let $f(x)=\cos (x')$ and $g(x)={{x}^{2}}$ then $\cos \left( {{x}^{2}} \right)=f\left( g\left( x \right) \right)$
$'g'$ is a function within $'f'$, so you call $'g'$ inner function and $'f'$ outer function. On the outer hand, $\cos (x).{{x}^{2}}$ is not a composite function. It is a product of $f(x)=\cos (x)$ and $g(x)={{x}^{2}}$ but neither function is within the other one. Usually, The only way to differentiate a composite function is to recognize that a function is composite and that the chain rule must be applied, You will not be able to differentiate correctly.

Note: Apply chain rule in given function. In some cases if you recognize composite functions you may get the inner and outer functions wrong. This will give you a derivative. For example, in the composite function ${{\cos }^{2}}(x)$ the outer function is ${{x}^{2}}$ and the inner function is $\cos (x)$ the outer function is ${{x}^{2}}$ and the inner function is $\cos (x)$ Sometimes you may get confused by this type of question and think $\cos (x)$ is the outer function.