
Find \[{{f}^{1}}\left( x \right)\].
\[f(x)=\sin \left( \dfrac{1}{{{x}^{2}}} \right)\]
Answer
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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.
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.
\[f(x)=\sin \left( \dfrac{1}{{{x}^{2}}} \right)\]. . . . . . . . . . . . . . . . . . . . . (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.
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\],\[\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)=\dfrac{-2}{{{x}^{3}}}\]. We can see one function is inside another we have to find internal derivatives.
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.
\[f(x)=\sin \left( \dfrac{1}{{{x}^{2}}} \right)\]. . . . . . . . . . . . . . . . . . . . . (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.
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