Question

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

Answer 1

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.