
Find \[{{f}^{1}}\left( x \right)\].
\[f\left( x \right)={{\cos }^{2}}\left( 3\sqrt{x} \right)\]
Answer
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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,
\[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\], \[\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}\]. 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,
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