Differentiation is a method to find rates of change and is an important topic for the JEE exam. The derivative of a function $y = f(x)$ of a variable $x$ is the rate of change of $y$ with respect to the rate of change of $x$. This article helps you to learn about the derivative of a function, standard derivatives, theorems of derivatives, differentiation of implicit functions, and higher order derivatives, all accompanied by solved examples.

Related Topics:

Limits, Continuity and Differentiability

Applications of Derivatives

Differential Equations

How to Differentiate a Function

The differentiation of a function is a way to calculate the rate of change of a function at a given point. For real-valued functions, it is the slope of the tangent line at a point on a graph.

The derivative of $y$ with respect to $x$ is often written as $\frac{\text{d}y}{\text{d}x}$ and is defined as the limit of the change in $y$ over the change in $x$ as the distance between $x_0$ and $x_1$ approaches zero, or infinitesimally.

In mathematical terms,

The derivative of $f$ at $a$, denoted $f’(a)$, is defined as: $$f’(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

( $$\begin{array}{l}{\displaystyle f’(a)=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}}\end{array}$$ )

Standard Derivatives

( $$\begin{array}{l}1]\ \frac{d}{dx}\left(u^n\right)=nu^{n-1}\frac{du}{dx}\end{array}$$ )

(\frac{d}{dx}\left(c\right)=0,; \text{where c is a constant})

( $$\begin{array}{l}\frac{dy}{dx}=\frac{\frac{\partial y}{\partial u}}{\frac{\partial u}{\partial x}} \text{ where } y=F(u) \text{ and } u=f(x)\end{array}$$ )

Chain Rule (or Function of a Function Rule)

4. Derivatives of Trigonometric Functions

( $$\begin{array}{l} \frac{d}{dx}\left(\sin u\right)=\cos u\frac{du}{dx}\ \frac{d}{dx}\left(\cos u\right)=-\sin u\frac{du}{dx}\ \frac{d}{dx}\left(\tan u\right)=\sec^2 u\frac{du}{dx}\ \frac{d}{dx}\left(\sec u\right)=\sec u\tan u\frac{du}{dx}\ \frac{d}{dx}(cosec\ u)=-cosecu\ cotu\ \frac{du}{dx}\ \frac{d}{dx}(cot\ u)=-cosec^2u\ \frac{du}{dx}\ \end{array}$$ )

5. Derivatives of Inverse Trigonometric Functions

( $$\begin{array}{l} \frac{d}{dx}\left(\sin^{-1} u\right) = \frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\qquad -1 < u < 1 \ \frac{d}{dx}\left(\cos^{-1} u\right) = \frac{-1}{\sqrt{1-u^2}}\frac{du}{dx},\qquad -1 < u < 1 \ \frac{d}{dx}\left(\tan^{-1} u\right) = \frac{1}{1+u^2}\frac{du}{dx} \ \frac{d}{dx}\ cosec^{-1}u = -\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx}\qquad |u| > 1 \ \frac{d}{dx}\left(\sec^{-1} u\right) = \frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx}\qquad |u| > 1 \ \frac{d}{dx}\left(\cot^{-1} u\right) = -\frac{1}{1+u^2}\frac{du}{dx} \ \end{array}$$ )

6. Exponential and Logarithmic Functions

( $$\begin{array}{l} \frac{d}{dx}\left(e^u\right)=e^u\frac{du}{dx}\ \frac{d}{dx}\left(a^u\right)=a^u\ln a\frac{du}{dx}, \text{where }a>0, a\ne1\ \frac{d}{dx}\left(\ln u\right)=\frac{1}{u}\frac{du}{dx}\ \frac{d}{dx}\left(\ln_a u\right)=\frac{1}{u\ln a}\frac{du}{dx},\text{ where }a>0, a\ne 1\ \end{array}$$ )

Hyperbolic Functions

**(\begin{array}{l} \frac{d}{dx}\left(\sinh u\right)=\cosh u\frac{du}{dx}\ \frac{d}{dx}\left(\cosh u\right)=\sinh u\frac{du}{dx}\ \frac{d}{dx}\left(\tanh u\right)=sech^{2}u\frac{du}{dx}\ \frac{d}{dx}\ (sech u)=-sechu\ tanhu\frac{du}{dx}\ \frac{d}{dx}\ (cosech u)=-cosechu\ cothu\frac{du}{dx}\ \frac{d}{dx}\ (coth u)=-cosech^{2}u\frac{du}{dx}\\end{array})

Inverse Hyperbolic Functions :eight:

$$\begin{array}{l} \frac{d}{dx}\sinh^{-1}u = \frac{1}{\sqrt{1 + u^2}}\frac{du}{dx} \\ \frac{d}{dx}\cosh^{-1}u = \frac{1}{\sqrt{u^2 - 1}}\frac{du}{dx}, \quad u > 1 \\ \frac{d}{dx}\tanh^{-1}u = \frac{1}{1 - u^2}\frac{du}{dx}, \quad |u| < 1 \\ \frac{d}{dx}\cosech^{-1}u = -\frac{1}{|u|\sqrt{u^2 + 1}}\frac{du}{dx}, \quad u \ne 0 \\ \frac{d}{dx}\sech^{-1}u = -\frac{1}{u\sqrt{1 - u^2}}\frac{du}{dx}, \quad 0 < u < 1 \\ \frac{d}{dx}\coth^{-1}u = \frac{1}{1 - u^2}\frac{du}{dx}, \quad |u| > 1 \end{array}$$

Some Standard Substitution

Expression Substitution

Simple tricks to solve complicated differential equations are listed below:

If function contains $\sqrt{a^2 - x^2}$; then substitute $x = a \sin\theta$ or $x = a \cos\theta$

(ii) If function contains $\sqrt{{{a}^{2}}+{{x}^{2}}}$; then substitute $x = a \cot{\theta}$ or $x = a \tan{\theta}$

(iii) If function contains $\sqrt{{{a}^{2}}+{{a}^{2}}\sec^2\theta}$ or $\sqrt{{{a}^{2}}+{{a}^{2}}\csc^2\theta}$ ; then substitute $x = a \cosec\theta$ or $x = a \sec \theta$

Theorems of Derivatives

Find below some of the important theorem results:

( $$\begin{array}{l} \frac{d}{dx}[u\pm v] = \frac{du}{dx} \pm \frac{dv}{dx} \ \frac{d}{dx}uv = u\frac{dv}{dx} + v\frac{du}{dx} \qquad\text{(Product Rule)}\ \frac{d}{dx}\frac{u}{v} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \qquad\text{(Quotient Rule)}\ \frac{d}{dx}\left( k,f(x) \right) = k\frac{d}{dx}(f(x)),,,\text{where},k,\text{is constant} \end{array}$$ )

Answer:

$$\frac{dy}{dx} = x \cos x \log x + \sin x \left(1 + \frac{\log x}{x}\right)$$

Solution: $$\begin{array}{l}{y^{’}}=1\times \sin x\log x+x\cos x\log x+x\sin x\times \frac{1}{x}=\sin x\log x+x\cos x\log x+\sin x\end{array}$$

Example 2: Find $\frac{dy}{dx}$ for $y = \sin(x^2 + 1)$.

Solution: $$y’ = \cos(x^2 + 1) \times 2x$$

2*cos(x^2 + 1)

Differentiation of Implicit Functions

In Implicit Differentiation, also known as the Chain Rule, differentiate both sides of an equation with two given variables by considering one of the variables as a function of the second variable. In other words, differentiate the given function with respect to x and solve for dy/dx. Let us take a look at some examples.

Answer:

Example 1: Find $\frac{dy}{dx}$ if $x^2 + 2xy + y^3 = 4$.

Solution: Differentiating both sides w.r.t. x, we get $$\frac{d}{dx}({{x}^{2}})+2\frac{d}{dx}(xy)+\frac{d}{dx}({{y}^{3}})=\frac{d}{dx}(4)$$

( $$\begin{array}{l}2x\frac{dy}{dx} + 2x\frac{dy}{dx} + 2y + 3y^2\frac{dy}{dx} = 0\end{array}$$ )

(\frac{d}{dx}\left(y\right)=-\frac{2\left(x+y\right)}{2x+3y^2})

Answer:

$$\frac{d}{d\sqrt{\cos x}}\log\sin x = \frac{\cos x}{\sin x\sqrt{\cos x}}$$

Solution: Let u = log sin x and v = (\sqrt{\cos x})

( $$\begin{array}{l} \frac{du}{dx} = \cot x \ \frac{dv}{dx} = \frac{-\sin x}{2\sqrt{\cos x}} \end{array}$$ )

(\frac{du}{dx}=-2\sqrt{\cos x}\cot x\csc x)

Higher Order Derivatives

Differentiation process can be continued up to the n^th derivative of a function. Generally, we deal with the first-order and second-order derivatives of the functions.

\frac{dy}{dx} is the first derivative of $y$ with respect to $x$

The second derivative of y with respect to x is d2y/dx2.

Similarly, finding the third, fourth, fifth and higher-order derivatives of any function, say g(x), are known as successive derivatives of g(x).

gn(x) or d^n/dx^n

Proof: Let (u=\tan^{-1} x)

Then (y=e^u)

(\frac{d}{dx} u = \frac{1}{1+x^2})

(\frac{dy}{dx} = \frac{d}{dx} e^u = e^u \frac{d}{dx} u = e^u \frac{1}{1+x^2})

(\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(e^u \frac{1}{1+x^2}\right) = e^u \frac{d}{dx} \frac{1}{1+x^2} + \frac{1}{1+x^2} \frac{d}{dx} e^u)

(\frac{d^2 y}{dx^2} = e^u \frac{-2x}{(1+x^2)^2} + \frac{1}{1+x^2} e^u \frac{1}{1+x^2})

(\frac{d^2 y}{dx^2} = e^u \frac{1-2x}{(1+x^2)^2})

(\frac{d^2 y}{dx^2} = (1-2x) \frac{dy}{dx})

Given:

This is a heading

Solution:

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( $$\begin{array}{l}{e}^{{{\tan }^{-1}}x} = y\end{array}$$ )

$$\frac{dy}{dx} = e^{\tan^{-1}x}\frac{1}{1+x^2}$$

(\frac{e^{\tan^{-1}x}}{1+x^{2}}) (i)

(\frac{d^2y}{dx^2} = \frac{(1+x^2)e^{\tan^{-1}x}\frac{1}{1+x^2} - 2xe^{\tan^{-1}x}}{(1+x^2)^2})

$$\frac{(1-2x)e^{\tan^{-1}x}}{(1+x^{2})^{2}}$$

(\frac{d^2y}{dx^2} \left(1 + x^2\right) = \frac{(1-2x)e^{\tan^{-1}x}}{\left(1+x^{2}\right)})

\frac{d}{dx}\left( (1-2x)y \right) (from eqn (i))

Therefore, it is proven.

Video Lessons

Methods of Differentiation – JEE Solved Questions

Important Theorems of Differentiation for JEE

Frequently Asked Questions

Differentiation in Mathematics refers to the process of finding the rate of change of a function with respect to one of its variables.

Differentiation is the process of finding the derivative of a function.

The product rule of differentiation states that if $f(x)$ and $g(x)$ are differentiable functions, then $(f(x)g(x))’ = f’(x)g(x) + f(x)g’(x)$.

Product Rule: $(\frac{d}{dx})(uv) = u(\frac{dv}{dx}) + v(\frac{du}{dx})$

The quotient rule of differentiation states that: $$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f’(x)g(x) - g’(x)f(x)}{[g(x)]^2}$$

(d/dx)(u/v) = (v (du/dx) - u (dv/dx))/v^2

What is the derivative of cot(x) with respect to x?

The derivative of cot x with respect to x = -cosec²x.