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Question: Find \[{{f}^{1}}\left( x \right)\]. \[f(x)=\sin \left( \dfrac{1}{{{x}^{2}}} \right)\]...

Find f1(x){{f}^{1}}\left( x \right).
f(x)=sin(1x2)f(x)=\sin \left( \dfrac{1}{{{x}^{2}}} \right)

Explanation

Solution

Hint : To solve the above problem we have to know the basic derivatives of sinx\sin xand 1x2\dfrac{1}{{{x}^{2}}}. After writing the derivatives rewrite the equation with the derivatives of the function.
ddx(sinx)=cosx\dfrac{d}{dx}\left( \sin x \right)=\cos x,ddx(1x2)=2x3\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)=\dfrac{-2}{{{x}^{3}}}. We can see one function is inside another we have to find internal derivatives.

Complete step-by-step answer :
The composite function rule shows us a quicker way. If f(x) = h(g(x)) then f (x) = h (g(x)) × g (x). In words: differentiate the 'outside' function, and then multiply by the derivative of the 'inside' function. ... The composite function rule tells us that f (x) = 17(x2 + 1)16 × 2x.

f(x)=sin(1x2)f(x)=\sin \left( \dfrac{1}{{{x}^{2}}} \right). . . . . . . . . . . . . . . . . . . . . (a)
ddx(sinx)=cosx\dfrac{d}{dx}\left( \sin x \right)=\cos x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
ddx(1x2)=2x3\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)=\dfrac{-2}{{{x}^{3}}}. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substituting (1) and (2) in (a) we get,
Therefore derivative of the given function is,
f1(x)=ddx(sin(1x2)){{f}^{1}}\left( x \right)=\dfrac{d}{dx}\left( \sin \left( \dfrac{1}{{{x}^{2}}} \right) \right)
We know the derivative of sinx\sin xand 1x2\dfrac{1}{{{x}^{2}}}. By writing the derivatives we get,
Further solving we get the derivative of the function as
f1(x)=cos(1x2)(2x3){{f}^{1}}\left( x \right)=\cos \left( \dfrac{1}{{{x}^{2}}} \right)\left( \dfrac{-2}{{{x}^{3}}} \right) . . . . . . . . . . . . . . . . . . . (3)
By solving we get,
f1(x)=2x3cos(1x2){{f}^{1}}\left( x \right)=\dfrac{-2}{{{x}^{3}}}\cos \left( \dfrac{1}{{{x}^{2}}} \right)

Note : In the above problem we have solved the derivative of the trigonometric function. In (3) the formation of 2x3\dfrac{-2}{{{x}^{3}}}is due to function in a function. In this case we have to find an internal derivative. Further solving for dydx\dfrac{dy}{dx}made us towards a solution. If we are doing derivative means we are finding the slope of a function. Care should be taken while doing calculations.