Question

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

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

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.

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

Note: In the above problem we have solved the derivative of the trigonometric function. In (3) the formation of $\dfrac{-2}{{{x}^{3}}}$is due to function in a function. In this case we have to find an internal derivative. Further solving for $\dfrac{dy}{dx}$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.