Differentiation of Implicit Function: Definition, Formula, Questions

Differentiation is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of differentiation have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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1. Implicit Function
2. Differentiation of Implicit Function
3. Solved Examples Based on Differentiation of Implicit Function
4. Summary

In this article, we will cover the concept of the Differentiation of Imlicit Function. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last five years of the JEE Main exam (from 2013 to 2023), a total of three questions have been asked on this concept, including one in 2021, one in 2022, and one in 2023.

Implicit Function

An implicit function is a function that includes both dependent and independent variables such as $F(x, y)=0$. For example, the equation of a circle $x^2+y^2=r^2$ is an implicit function because $y$ is not explicitly expressed as a function of $x$.

Differentiation of Implicit Function

Implicit differentiation means differentiation on both sides of the equation concerning the independent variable with the help of the Chain rule.

Chain Rule:
If $y=u o v(x)=u[v(x)]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} u\{v(x)\}}{\mathrm{d}\{v(x)\}} \times \frac{\mathrm{d}}{\mathrm{d} x} v(x)
$

is known as the chain rule. Or,

If $y=f(u)$ and $u=g(x)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} u} \cdot \frac{\mathrm{d} u}{\mathrm{~d} x}$
The chain rule can be extended as follows:

$
\begin{aligned}
& \text { If } y=[\operatorname{uovow}(x)]=u[v\{w(x)\}] \text {, then } \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}[u[v\{w(x)\}]}{\mathrm{d} v\{w(x)\}} \times \frac{\mathrm{d}[v\{w(x)\}]}{\mathrm{d} w(x)} \times \frac{\mathrm{d}[w(x)]}{\mathrm{d} x}
\end{aligned}
$
If variables $x$ and $y$ are connected by a relation of the form $f(x, y)=0$ and it is not possible or convenient to express $y$ as a function of $x$ i.e., in the form $\mathrm{y}=\Phi(\mathrm{x})$, then $y$ is said to be an implicit function.

To find $\frac{d y}{d x}$ in such a case, we differentiate both sides of the given relation concerning $x$ keeping in mind that the derivative of $\Phi(\mathrm{y})$ concerning $x$ is $\frac{d \phi}{d y} \times \frac{d y}{d x}$

For example

$
\frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}, \frac{d}{d x}\left(y^2\right)=2 y \frac{d y}{d x}
$
It should be noted that $\frac{d}{d y}(\sin y)=\cos y$ but $\frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}$

Recommended Video Based on Differentiation of Implicit Function

Solved Examples Based on Differentiation of Implicit Function

Example 1: If $y=y(x)$ is an implicit function of x such that $\log _e(x+y)=4 x y$, then $\frac{d^2 y}{d x^2}$ at $x=0$ is equal to $\qquad$
[JEE Main 2021]
1) $40$
2) $3$
3) $2$
4) $1$

Solution:

$
\begin{aligned}
& \ln (x+y)=4 x y \\
& \frac{1}{x+y}\left(1+\frac{d y}{d x}\right)=4 x \frac{d y}{d x}+4 y---(i)
\end{aligned}
$
At $x=0$

$
\begin{aligned}
& \ln (y)=0 \Rightarrow y=1 \\
& \& \frac{1}{0+1}\left(1+\frac{d y}{d x}\right)=0+4 \Rightarrow \frac{d y}{d x}=3
\end{aligned}
$
From (i)

$
1+\frac{d y}{d x}=(x+y)\left(4 x \frac{d y}{d x}+4 y\right)
$
Differentiating

$
\begin{aligned}
& \frac{d^2 y}{d x^2}=(x+y)\left(4 x \frac{d^2 y}{d x^2}+4 \frac{d y}{d x}+4 \frac{d y}{d x}\right)+\left(1+\frac{d y}{d x}\right)\left(4 x \frac{d y}{d x}+4 y\right) \\
& \text { Put } x=0, y=1, \frac{d y}{d x}=3 \\
& \Rightarrow \frac{d^2 y}{d x^2}=1(0+12+12)+4 \cdot(0+4) \\
& \Rightarrow \frac{d^2 y}{d x^2}=24+16=40
\end{aligned}
$

Hence, the answer is the option 1.

Example 2: For the curve $\mathrm{C}:\left(\mathrm{x}^2+\mathrm{y}^2-3\right)+\left(\mathrm{x}^2-\mathrm{y}^2-1\right)^5=0$, the value of $3 y^{\prime}-y^3 y^{\prime \prime}$, at the point $(\alpha, \alpha), \alpha>0$, on C , is equal to $\qquad$
[JEE Main 2022]
1) $16$
2) $3$
3) $1$
4) $0$

Solution:

$
\begin{aligned}
& \left(\alpha^2+\alpha^2-3\right)+\left(\alpha^2-\alpha^2-1\right)^5=1 \\
& \Rightarrow 2 \alpha^2=4
\end{aligned}
$

$(\alpha, \alpha)$ lies on given curve, $\operatorname{so} \Rightarrow \alpha=\sqrt{2} \quad$ (as $\alpha>0$ )
Differentiating the curve

$
2 \mathrm{x}+2 \mathrm{yy}^{\prime}+5\left(\mathrm{x}^2-\mathrm{y}^2-1\right)^4\left(2 \mathrm{x}-2 \mathrm{y} y^{\prime}\right)=0
$
Put $\mathrm{x}=\mathrm{y}=\sqrt{2}$.

$
\begin{aligned}
& 2 \sqrt{2}+2 \sqrt{2} y^{\prime}+5 \cdot\left(2 \sqrt{2}-2 \sqrt{2} y^{\prime}\right)=0 \\
& \Rightarrow 12 \sqrt{2}-8 \sqrt{2} y^{\prime}=0 \\
& \Rightarrow y^{\prime}=\frac{3}{2}
\end{aligned}
$

Differentiating again

$
2+2\left(\mathrm{yy}^{\prime \prime}+\left(\mathrm{y}^{\prime}\right)^2\right)+20\left(\mathrm{x}^2-\mathrm{y}^2-1\right)\left(2 \mathrm{x}-2 \mathrm{yy} y^{\prime}\right)\left(2 \mathrm{x}-2 \mathrm{yy} y^{\prime}\right)+5\left(\mathrm{x}^2-\mathrm{y}^2-1\right)^4
$
Put $x=y=\sqrt{2}$ and $y^{\prime}=\frac{3}{2}$

$
\begin{aligned}
& y^{\prime \prime}=-\frac{23}{4 \sqrt{2}} \\
& \therefore \quad 3 y^{\prime}-y^3 y^{\prime \prime}=3 \times \frac{3}{2}-(2 \sqrt{2}) \times\left(-\frac{23}{4 \sqrt{2}}\right) \\
& =\frac{9}{2}+\frac{23}{2}=16 \\
\end{aligned}
$
Hence, the answer is 16.

Example 3: If $2 x^y+3 y^x=20$, then $\frac{d y}{d x}$ at $(2,2)$ is equal to :
[JEE Main 2023]
1) $-\left(\frac{3+\log _e 8}{2+\log _e 4}\right)$
2) $-\left(\frac{2+\log _e 8}{3+\log _e 4}\right)$
3) $-\left(\frac{3+\log _e 4}{2+\log _e 8}\right)$
4) $-\left(\frac{3+\log _e 16}{4+\log _e 8}\right)$

Solution:

$\begin{aligned} & 2 x^y+3 y^x=20 \\ & v_1^{v_2}\left(v_2 \frac{1}{v_1}+\ln 1 \cdot v_2^1\right) \\ & 2 x^y\left(y \cdot \frac{1}{x}+\ln x \frac{d y}{d x}\right)+3 y^x\left(x \frac{1}{y} \cdot \frac{d y}{d x}+\ln y \cdot 1\right)=0 \\ & \operatorname{Put}(2,2) \\ & 2.4\left(1+\ln 2 \frac{d y}{d x}\right)+3 \cdot 4\left(1 \cdot \frac{d y}{d x}+\ln 2\right)=0 \\ & \frac{d y}{d x}[8 \ln 2+12]+8+12 \ln 2=0 \\ & \frac{d y}{d x}=-\left[\frac{2+3 \ln 2}{3+2 \ln 2}\right]=-\left[\frac{2+\ln 8}{3+\ln 4}\right]\end{aligned}$

Hence, the answer is the option 2.

Example 4: An implicit function $y$ is such that $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sin x}{y^2}$, then $\frac{\mathrm{d} \sqrt{y}}{\mathrm{~d} x}=$
1) $\frac{1}{2}(\sin x) \cdot y^{\frac{-x}{2}}$
2) $(\sin x) \cdot y^{\frac{3}{2}}$
3) $(\sin x) \cdot \sqrt{y}$
4) $(\sin x) \cdot y^{\frac{-5}{2}}$

Solution:
As we have learnt,
Derivative of implicit function -

When y is not expressible explicitly in terms of x then we take, a derivative of $y^n$ as

$
n y^{n-1} \times \frac{d y}{d x}
$
For example

$
\begin{aligned}
& \frac{d}{d x}\left(y^2\right)=2 y \cdot \frac{d y}{d x} \\
& \frac{d}{d x}\left(y^3\right)=3 y^2 \cdot \frac{d y}{d x} \\
& \frac{\mathrm{d}}{\mathrm{d} x} \sqrt{y}=\frac{\mathrm{d}}{\mathrm{d} x} y^{\frac{1}{2}}=\frac{1}{2}(y)^{-\frac{1}{2}} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2 \sqrt{y}} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x} \\
& \Rightarrow \frac{\mathrm{d}}{\mathrm{d} x} \sqrt{y}=\frac{1}{2}(y)^{-\frac{1}{2}} \cdot \frac{\sin x}{y^2}=\frac{1}{2}(y)^{-\frac{5}{2}} \cdot \sin x
\end{aligned}
$

Hence, the answer is the option 1.

Example 5: Let $y=\sin (x+y)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$
1) $\frac{\cos (x+y)}{1+\cos (x+y)}$
2) $\frac{\cos (x+y)}{\cos (x+y)-1}$
3) $\frac{\cos (x+y)}{1-\cos (x+y)}$
4) $-\frac{\cos (x+y)}{1+\cos (x+y)}$

Solution:
As we have learned,
Derivative of implicit function -
When $y$ is given in any function then we find the derivative of the function first and then find the derivative of $y$ and collect the terms containing
$\frac{dy}{dx}$ on the left side and find $\frac{dy}{dx}$ in terms of $x$ and $y$
Now Differentiating both sides we have,

$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=(\cos (x+y))\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right) \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos (x+y)+\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \cos (x+y) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}(1-\cos (x+y))=\cos (x+y) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\cos (x+y)}{1-\cos (x+y)}
\end{aligned}
$

Hence, the answer is the option 3.

Summary

Implicit differentiation is a crucial tool in calculus that helps us differentiate functions not explicitly defined. By using the chain rule and solving for the needed derivative, implicit differentiation lets us analyze complex relationships between variables. Its uses are widespread, from physics and engineering to economics and biology, showcasing its importance and adaptability.