Derivative of a Function wrt Another Function

Function

If $A$ and $B$ are two non-empty sets, then a relation from $A$ to $B$ is said to be a function if each element $x$ in $A$ is assigned a unique element $f(x)$ in $B$.
It is written as $f: A \rightarrow B$ and read as “$f$ is a mapping from $A$ to $B$.”

Derivative of a Function

Let $f$ be defined on an open interval $I \subseteq \mathbb{R}$ containing the point $x_0$, and suppose that
$\lim_{\Delta x \rightarrow 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$ exists.
Then $f$ is said to be differentiable at $x_0$, and the derivative of $f$ at $x_0$, denoted by $f'(x_0)$, is given by:

$f'(x_0) = \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$

For all $x$ for which this limit exists,

$f'(x) = \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$

is a function of $x$.

In addition to $f'(x)$, other notations used to represent the derivative of $y = f(x)$ include
$f'(x)$, $\frac{dy}{dx}$, $y'$, $\frac{d}{dx}[f(x)]$, $D_x[y]$, $Dy$, or $y_1$.
Here, $\frac{d}{dx}$ or $D$ is known as the differential operator.

Properties of the Derivative of a Function

We have provided below the properties and rules of differentiation of a function:

1. Sum Rule:
   The derivative of the sum of two functions equals the sum of their derivatives.
   $\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

2. Difference Rule:
   The derivative of the difference of two functions equals the difference of their derivatives.
   $\frac{d}{dx}[f(x) - g(x)] = \frac{d}{dx}[f(x)] - \frac{d}{dx}[g(x)]$

3. Product Rule:
   The derivative of the product of two functions is given by:
   $\frac{d}{dx}[f(x)g(x)] = \left(\frac{d}{dx}f(x)\right)g(x) + f(x)\left(\frac{d}{dx}g(x)\right)$

4. Quotient Rule:
   The derivative of the quotient of two functions is given by:
   $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)\left(\frac{d}{dx}f(x)\right) - f(x)\left(\frac{d}{dx}g(x)\right)}{[g(x)]^2}$

Derivative of a Function with respect to Another Function

The derivative of a function with respect to another function refers to differentiating two interdependent functions.
To find this, we use the Chain Rule, which allows us to differentiate one function in terms of another.

Chain Rule

Let $f(x)$ and $g(x)$ be two functions. To differentiate $f(x)$ with respect to $g(x)$, let:
$u = f(x)$ and $v = g(x)$.

Then,

$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}$

where $\frac{dv}{dx} \neq 0$.

Relation Between $\frac{dy}{dx}$, $\frac{dy}{du}$ and $\frac{du}{dx}$

When two or more functions are related through a common variable, we often need to find the derivative of one function with respect to another. This connection is beautifully explained through the chain rule of differentiation, which helps us express $\frac{dy}{dx}$ in terms of $\frac{dy}{du}$ and $\frac{du}{dx}$.

In simple words, if one variable depends on another, and that second variable depends on a third, then their rates of change are connected through multiplication.

Derivation of $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Let $y$ be a function of $u$, and $u$ be a function of $x$, such that:

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

Then, by the chain rule, the derivative of $y$ with respect to $x$ is obtained as follows:

$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
$

This means the rate of change of $y$ with respect to $x$ is equal to the rate of change of $y$ with respect to $u$ multiplied by the rate of change of $u$ with respect to $x$.

Step-by-Step Derivation:

1. Since $y = f(u)$, a small change in $u$ causes a small change in $y$.
   So, $\Delta y = f(u + \Delta u) - f(u)$

2. Similarly, since $u = g(x)$, a small change in $x$ leads to a change in $u$:
   $\Delta u = g(x + \Delta x) - g(x)$

3. Dividing both sides, we can write:
   $
   \frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}
   $

4. Taking the limit as $\Delta x \rightarrow 0$, we get:
   $
   \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
   $

Hence proved.

Example: Let $y = \sin(u)$ and $u = 3x^2$.
Then, $
\frac{dy}{du} = \cos(u)
$ and $
\frac{du}{dx} = 6x
$

By the chain rule,
$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \times 6x = 6x \cos(3x^2)
$

This example shows how chain differentiation simplifies complex compositions of functions into smaller, easier parts.

Important Formulae for Derivative of a Function with Respect to Another Function

Below is a quick reference table of the key formulas used when differentiating one function with respect to another. These relationships make it easier to apply the chain rule and handle composite or interconnected functions smoothly.

Meaning of $\dfrac{dy}{dx}$ When Both $y$ and $x$ Are Functions of $t$

In many Class 12 calculus problems, $y$ and $x$ are expressed in terms of a third variable $t$, called a parameter.

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

Here, $y$ does not depend on $x$ directly. The derivative $\dfrac{dy}{dx}$ represents the rate of change of $y$ with respect to $x$, calculated using the common parameter $t$.

The idea is:

• First find how fast $y$ changes with $t$

• Then find how fast $x$ changes with $t$

• Finally relate both rates

This concept is known as derivative of a function with respect to another function and is a key part of parametric differentiation in Class 12.

Formula for Derivative of a Function with Respect to Another Function

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

then the derivative of $y$ with respect to $x$ is given by:

$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

This formula is fundamental in:

• Parametric equations

• Curve tracing

• Applications of derivatives

• Board and competitive exam questions

Step-by-Step Method to Find $\dfrac{dy}{dx}$ Using Parametric Differentiation

To find $\dfrac{dy}{dx}$, follow this standard exam-oriented method:

Step 1: Differentiate $y$ with respect to $t$ to get
$\dfrac{dy}{dt}$

Step 2: Differentiate $x$ with respect to $t$ to get
$\dfrac{dx}{dt}$

Step 3: Divide the two derivatives

$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

This method is simple, systematic, and widely used in Class 12 parametric differentiation problems.

Important Conditions for Differentiating One Function with Respect to Another

Before applying the formula, keep these important conditions in mind:

• $\dfrac{dx}{dt} \neq 0$, otherwise $\dfrac{dy}{dx}$ is not defined

• Both $x$ and $y$ must be differentiable functions of $t$

• The parameter $t$ should lie in the valid domain of both functions

Following these conditions helps avoid calculation errors and ensures correct answers in board examinations and entrance tests.

Solved Examples Based on Derivative of a Function with Respect to Another Function

Example 1:Find the second order derivative if x and y are given by $x=a \sin t$ and $y=a \operatorname{cost}$
1) $
\frac{1}{a} \cos ^3 t
$

2) $
-\frac{1}{a} \cos ^3 t
$

3) $
\frac{1}{a} \sec ^3 t
$

4) $
-\frac{1}{a} \sec ^3 t
$

Solution:

Differentiating the function implicitly with respect to " $x$ ";

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{d y}{d t}}{\frac{\mathrm{~d} x}{d t}}=\frac{-a \sin t}{a \cos t}=-\frac{\sin t}{\cos t}
$

Again differentiating with respect to " $x$ ";

$
\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{\mathrm{d}}{\mathrm{~d} x}\left(\frac{d y}{d x}\right)=\frac{\mathrm{d}}{\mathrm{~d} t}\left(-\frac{\sin t}{\cos t}\right) \frac{d t}{d x} \\
& \frac{d^2 y}{d x^2}=-\left(\sec ^2 t\right) \times \frac{1}{a \cos t}=-\frac{\sec ^2 t}{a \cos t}=-\frac{1}{a} \sec ^3 t
\end{aligned}
$

Hence, the answer is the option 4.

Example 2: Differentiate $\cos \left(a x^2+b x+c\right)$ with respect to $sin$ $\left(l x^2+m x+n\right)$
1) $
\frac{-(2 a x+b) \cos \left(a x^2+b x+c\right)}{(2 l x+m) \sin \left(l x^2+m x+n\right)}
$

2) $
\frac{-(2 a x+b) \sin \left(a x^2+b x+c\right)}{(2 l x+m) \cos \left(l x^2+m x+n\right)}
$

3) $
\frac{-(2 l x+m) \cos \left(l x^2+m x+n\right)}{(2 a x+b) \sin \left(a x^2+b x+c\right)}
$

4) $
\frac{-(2 l x+m) \sin \left(l x^2+m x+n\right)}{(2 a x+b) \cos \left(a x^2+b x+c\right)}
$

Solution:

Let $f(x)=\cos \left(a x^2+b x+c\right)$ and $g(x)=\sin \left(l x^2+m x+n\right)$

$
\begin{aligned}
& \frac{d f}{d g}=\frac{f^{\prime}(x)}{g^{\prime}(x)} \\
& f^{\prime}(x)=-\sin \left(a x^2+b x+c\right)(2 a x+b) \\
& g^{\prime}(x)=\cos \left(l x^2+m x+n\right)(2 l x+m) \\
& \frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{-(2 a x+b) \sin \left(a x^2+b x+c\right)}{(2 l x+m) \cos \left(l x^2+m x+n\right)}
\end{aligned}
$

Hence, the answer is the option 2.

Example 3: Find the derivative of $\log _{10}(x)$ with respect to $x^3$.
1) $
\frac{1}{2 x^2\left(\log _e 10\right)}
$

2) $
\frac{1}{2 x^3\left(\log _e 10\right)}
$

3) $
\frac{1}{3 x^3\left(\log _e 10\right)}
$

4) $
\frac{1}{3 x^2\left(\log _e 10\right)}
$

Solution:

Let $u =\log _{10}(x)$ and $v=x^3$
Also,$\frac{d u}{d x}=\frac{1}{x \log _e(10)}$ and $\frac{d v}{d x}=3 x^2$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$
$= \frac{\frac{1}{x \log _c(10)}}{3 x^2}=\frac{1}{3 x^3 \log _e(10)}$

Hence, the answer is the option 3.

Example 4: If $y=\frac{\sqrt{\left(1+t^2\right)}-\sqrt{\left(1-t^2\right)}}{\sqrt{\left(1+t^2\right)}+\sqrt{\left(1-t^2\right)}}$ and $x=\sqrt{\left(1-t^4\right)}$ then $\frac{d y}{d x}$ is equal to
1) $
\frac{-1}{t^2\left\{1+\sqrt{\left.\left(1-t^4\right)\right\}}\right.}
$

2) $
\frac{\left\{\sqrt{\left(1-t^4\right)}-1\right\}}{t^6}
$

3) $
\frac{1}{t^2\left\{1+\sqrt{\left.\left(1-t^4\right)\right\}}\right.}
$

4) $
\frac{1-\sqrt{\left(1-t^4\right)}}{t^6}
$

Solution:
$
\begin{aligned}
& y=\frac{\left(\sqrt{1+t^2}-\sqrt{1-t^2}\right)^2}{\left(\sqrt{1+t^2}\right)^2-\left(\sqrt{1-t^2}\right)^2} \\
& =\frac{2-2 \sqrt{1-t^4}}{2 t^2}=\frac{1-\sqrt{1-t^4}}{t^2} \\
& \frac{d y}{d t}=\frac{t^2\left\{0-\frac{1}{2 \sqrt{1-t^4}} \times-4 t^3\right\}-\left\{1-\sqrt{1-t^4}\right\} 2 t}{t^4} \\
& =\frac{2 t^5}{\sqrt{1-t^4}}-2 t\left\{1-\sqrt{1-t^4}\right\} \\
& =\frac{2\left\{1-\sqrt{1-t^4}\right\}}{t^3 \sqrt{1-t^4}} \\
& \frac{d x}{d t}=\frac{-2 t^3}{\sqrt{1-t^4}} \\
& \frac{d y}{d x}=\frac{\sqrt{1-t^4}-1}{t^6}=\frac{-1}{t^2\left\{1+\sqrt{1-t^4}\right\}}
\end{aligned}
$
Hence, the answer is the option 1.

Example 5: If $y=x^3+e^x$, then find $\frac{d^2 y}{d x^2}$.
1) $
-\frac{\left(5 x+e^x\right)}{\left(3 x^2+e^x\right)^3}
$

2) $
-\frac{\left(6 x+e^x\right)}{\left(3 x^2+e^x\right)^3}
$

3) $
-\frac{\left(6 x+e^x\right)}{\left(3 x^2+e^x\right)^2}
$

4) $
-\frac{\left(5 x+e^x\right)}{\left(3 x^2+e^x\right)^2}
$

Solution:

Given that, $y=x^3+e^x$

$\begin{aligned} & \frac{d y}{d x}=3 x^2+e^x \\ & \Rightarrow \frac{d x}{d y}=\frac{1}{3 x^2+e^x} \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{1}{\left(3 x^2+e^x\right)^2} \frac{d}{\mathrm{~d} y}\left(3 x^2+e^x\right) \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{1}{\left(3 x^2+e^x\right)^2}\left[6 x+e^x\right] \frac{d x}{d y} \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{6 x+e^x}{\left(3 x^2+e^x\right)^2} \times \frac{1}{\left(3 x^2+e^x\right)} \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{6 x+e^x}{\left(3 x^2+e^x\right)^3}\end{aligned}$

Hence, the answer is the option (2)

List of topics related to the Derivative of a Function with respect to Another Function

This section covers key subtopics linked to the concept of the derivative of one function with respect to another, to learn in detail about them.

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

Here you’ll find NCERT-based resources such as exemplar problems, step-by-step solutions, and detailed notes designed to strengthen your understanding of the derivative of a function with respect to another function — perfect for Class 12 and JEE preparation.

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 Derivative of a Function with respect to Another Function

This section offers a curated set of solved and unsolved practice questions to help you master $\frac{dy}{du}$ and $\frac{du}{dx}$ relationships, improve problem-solving accuracy, and prepare effectively for board and competitive exams.

Differentiation Of A Function Wrt Another Function And Higher Order Derivative Of A Function- Practice Question MCQ

We have shared below the links to practice questions on the related topics of Derivative of a Function with respect to another function: