Question

How do you find \[\dfrac{dy}{dx}\]by implicit differentiation of\[x=\sec \left( \dfrac{1}{y} \right)\]?

Answer 1

Hint:
To find the \[\dfrac{dy}{dx}\] by implicit differentiation of \[x=\sec \left( \dfrac{1}{y} \right)\], we need to differentiate the given function using the chain rule of differentiation. We need to let \[\dfrac{1}{y}=u\]and substituting the value we will find the derivative. Later by applying the trigonometric relations i.e.\[\sec \theta =\dfrac{1}{\cos \theta }\ and\ \tan \theta =\dfrac{1}{\cot \theta }\], we need to multiply the given derivative by these trigonometric function. In this way we will get the required answer.

Formula used:
1) Chain rule of differentiation:
If \[f\] and \[g\] are both differentiable and \[F\left( x \right)\] is the composite function defined by \[F\left( x \right)=f\left( g\left( x \right) \right)\] then \[F\] is differentiable and \[F'\] is given by the product,\[F'\left( x \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)\]
2) Power rule of differentiation:
The functions of the form of \[f\left( x \right)={{x}^{n}}\], the power rule is used to derivative them i.e.,\[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\]
3) \[\sec \theta =\dfrac{1}{\cos \theta }\ and\ \tan \theta =\dfrac{1}{\cot \theta }\]

Complete step by step solution:
We have given that,
\[x=\sec \left( \dfrac{1}{y} \right)\]
Differentiate both the sides with respect to ‘x’,
\[\dfrac{d}{dx}x=\dfrac{d}{dx}\left( \sec \left( \dfrac{1}{y} \right) \right)\]
Simplifying the above, we will get
\[\dfrac{dx}{dx}=\dfrac{d}{dx}\left( \sec \left( \dfrac{1}{y} \right) \right)\]
Cancelling out the common terms, we will get
\[1=\dfrac{d}{dx}\left( \sec \left( \dfrac{1}{y} \right) \right)\]
Taking the derivative of both the sides using the chain rule of differentiation which is given by \[F'\left( x \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)\],
Let \[\dfrac{1}{y}=u\]
Substituting the values, we will get
\[1=\dfrac{d}{dx}\left( \sec \left( u \right) \right)\]
Applying the chain rule of differentiation,
\[1=\dfrac{d}{du}\sec u\cdot \dfrac{d}{dx}u\]
As we know that the sec rule of differentiation i.e. \[\dfrac{d}{dx}\sec x=\sec x\tan x\],
Thus,
\[1=\sec u\tan u\cdot \dfrac{d}{dx}u\]
Undo the substitution i.e. \[\dfrac{1}{y}=u\],
\[1=\sec \left( \dfrac{1}{y} \right)\tan \left( \dfrac{1}{y} \right)\cdot \dfrac{d}{dx}\left( \dfrac{1}{y} \right)\]
Rewritten the above as,
\[1=\sec \left( \dfrac{1}{y} \right)\tan \left( \dfrac{1}{y} \right)\cdot \dfrac{d}{dx}\left( {{y}^{-1}} \right)\]
Using the general power rule of differentiation i.e.\[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\]
Therefore,
\[1=\sec \left( \dfrac{1}{y} \right)\tan \left( \dfrac{1}{y} \right)\cdot \left( -{{y}^{-2}} \right)\dfrac{dy}{dx}\]
Multiplying both the sides by\[-{{y}^{2}}\], we will get
\[-{{y}^{2}}=\sec \left( \dfrac{1}{y} \right)\tan \left( \dfrac{1}{y} \right)\cdot \dfrac{dy}{dx}\]
Multiplying both the sides by \[\cos \left( \dfrac{1}{y} \right)\]and\[\cot \left( \dfrac{1}{y} \right)\],
As we know that the relation between the trigonometric function i.e.,
\[\sec \theta =\dfrac{1}{\cos \theta }\ and\ \tan \theta =\dfrac{1}{\cot \theta }\]
Thus, we will obtained
\[-{{y}^{2}}\cos \left( \dfrac{1}{y} \right)\cot \left( \dfrac{1}{y} \right)=\dfrac{dy}{dx}\]
Therefore,

\[\dfrac{dy}{dx}=-{{y}^{2}}\cos \left( \dfrac{1}{y} \right)\cot \left( \dfrac{1}{y} \right)\]
Hence, this is the required answer.

Note:
Students should remember that the function \[\sec \left( \dfrac{1}{y} \right)\]is a composite function. So it is of the form\[F\left( x \right)=f\left( g\left( x \right) \right)\], so to find its derivative we need to use the chain rule of differentiation i.e. given by\[F'\left( x \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)\]. Students make such mistakes and end up getting a wrong answer. Thus students should need to remember all the basic rules and formulas of differentiation to avoid such types of mistakes.