Question

Find ${{f}^{1}}\left( x \right)$.
$f\left( x \right)={{\cos }^{2}}\left( 3\sqrt{x} \right)$

Answer 1

Hint:To solve the above problem we have to know the basic derivatives of $\cos x$and $\sqrt{x}$. After writing the derivatives rewrite the equation with the derivatives of the function.
$\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}$. We can see one function is inside another we have to find internal derivatives.

Complete step-by-step answer:
Therefore derivative of the given function is,
${{f}^{1}}\left( x \right)=\dfrac{d}{dx}\left( {{\cos }^{2}}\left( 3\sqrt{x} \right) \right)$
We know the derivative of $\cos x$and $\sqrt{x}$ . By writing the derivatives we get,
Further solving we get the derivative of the function as
$=2\cos (3\sqrt{x})\dfrac{d}{dx}\left( \cos (3\sqrt{x}) \right)$. . . . . . . . . . . . . . . . . . . (3)
By solving we get, we are solving for internal derivative,
$=2\cos (3\sqrt{x})(-\sin \left( 3\sqrt{x} \right)\dfrac{d}{dx}\left( 3\sqrt{x} \right)$
$=2\cos (3\sqrt{x})(-\sin \left( 3\sqrt{x} \right)\left( \dfrac{3}{2\sqrt{x}} \right)$
$=-\dfrac{6}{2\sqrt{x}}\cos (3\sqrt{x})\sin (3\sqrt{x})$
The derivative of the given function is $=-\dfrac{3}{\sqrt{x}}\cos (3\sqrt{x})\sin (3\sqrt{x})$

Note: In the above problem we have solved the derivative of the trigonometric function. In (3) the formation of $\dfrac{1}{2\sqrt{x}}$is due to function in a function. In this case we have to find an internal derivative. If we are doing derivative means we are finding the slope of a function. Care should be taken while doing calculations.$f\left( x \right)={{\cos }^{2}}\left( 3\sqrt{x} \right)$. . . . . . . . . . . . . . . . . . . . . (a)
$\dfrac{d}{dx}\left( \cos x \right)=-\sin x$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
$\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substituting (1) and (2) in (a) we get,