Differentiation of Implicit Function: Definition, Formula, Questions

What is an Implicit Function?

An implicit function is a function in which the dependent variable $y$ is not explicitly expressed in terms of the independent variable $x$. Instead, the function is defined through an equation involving both $x$ and $y$, such as $F(x, y) = 0$. These functions are common in algebra, calculus, and physics because not all relationships can be neatly solved for $y$.

Example: The equation of a circle $x^2 + y^2 = r^2$ defines $y$ implicitly in terms of $x$, since $y$ can take two values for each $x$ within the domain $-r \leq x \leq r$.

Difference Between Explicit and Implicit Functions

Understanding the difference helps in selecting the right differentiation technique.

Explicit function: $y$ is expressed directly as a function of $x$.

Example: $y = 2x + 3$ or $y = x^2 + \sin x$

$ \frac{dy}{dx} $ can be found using standard differentiation rules.

Implicit function: $y$ is not isolated; it satisfies an equation $F(x, y) = 0$.

Example: $x^2 + y^2 = 25$ or $xy + \sin y = x^2$

$ \frac{dy}{dx} $ must be found using implicit differentiation.

Differentiability and Existence of Derivative in Implicit Functions

Differentiability determines whether the derivative $\frac{dy}{dx}$ exists for an implicit function $F(x, y) = 0$. A function is differentiable at a point if it can be locally expressed as a differentiable function of $x$ and if $\frac{\partial F}{\partial y} \neq 0$. Continuity of $F(x, y)$ is also necessary, because a discontinuous function cannot be differentiated.

Key conditions for differentiability:

• $F(x, y)$ is continuous near the point of interest.

• $\frac{\partial F}{\partial y} \neq 0$ at the point.

• The slope $\frac{dy}{dx}$ can be determined unambiguously.

Graphical examination helps identify points of non-differentiability, such as sharp corners or vertical tangents. For example, for the circle $x^2 + y^2 = r^2$,
$\frac{dy}{dx} = -\frac{x}{y}$
exists everywhere except at $y=0$, which corresponds to vertical tangents at the top and bottom points of the circle.

Basic Rules of Implicit Differentiation

Implicit differentiation is an important rule of differentiation that allows us to calculate $\frac{dy}{dx}$ without explicitly solving for $y$. The chain rule is the backbone of this method.

Using the chain rule:

• If $y$ appears as $y^n$, then
  $\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}$

• Example: For $x^2 + y^2 = 25$, differentiating both sides gives
  $\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$

Derivatives of standard implicit curves:

• Circle: $x^2 + y^2 = r^2 \implies \frac{dy}{dx} = -\frac{x}{y}$

• Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$

• Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \implies \frac{dy}{dx} = \frac{b^2 x}{a^2 y}$

Step-by-step method to find $\frac{dy}{dx}$:

1. Differentiate both sides of the equation w.r.t $x$, applying the chain rule for $y$.

2. Collect all terms containing $\frac{dy}{dx}$ on one side.

3. Solve for $\frac{dy}{dx}$.

4. Simplify the expression if possible.

Example: For $x^3 + y^3 = 6xy$, differentiating gives
$3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$

Differentiation of Parametric Forms

Parametric differentiation is a method used when both $x$ and $y$ are expressed as functions of a third variable, usually denoted $t$. This approach is useful for analyzing curves where $y$ cannot be expressed explicitly as a function of $x$. Many curves in physics, engineering, and higher mathematics, such as motion trajectories or cycloids, are naturally described in parametric form.

In parametric differentiation, we treat $x$ and $y$ as functions of a common parameter $t$:

$x = f(t)$ and $y = g(t)$

The goal is to find the derivative $ \frac{dy}{dx} $, which represents the slope of the curve in the $xy$-plane. Since $x$ and $y$ are both functions of $t$, we use the chain rule to relate their rates of change with respect to $t$.

The standard formula for parametric differentiation is:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Here, $ \frac{dy}{dt} $ is the derivative of $y$ with respect to $t$, and $ \frac{dx}{dt} $ is the derivative of $x$ with respect to $t$. This formula allows us to calculate the slope of the curve at any point $t$ without explicitly expressing $y$ in terms of $x$.

Key points:

The formula only works if $dx/dt \neq 0$.

Parametric differentiation is often combined with implicit differentiation for more complex curves.

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.

List of topics related to Differentiation of Implicit Functions

This section highlights all key topics for Class 12 implicit differentiation, helping students focus on essential concepts for board exams and competitive tests like JEE and CUET.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity and Discontinuity

Continuity of Composite Function

Continuity And Differentiability

Differentiability of Composite Function

NCERT Resources

This section provides reliable NCERT notes, solutions, and exemplar problems to strengthen Class 12 students’ understanding of implicit differentiation.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Differentiation of Implicit Functions

Test your skills with carefully designed MCQs and practice questions, improving accuracy and speed for board exams, JEE, and CUET.

Differentiation Of Implicit Function- Practice Question MCQ

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