Answer
Verified
396k+ views
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE