Differentiation

Applied Mathematics Class 11 - Chapter 8

1. Concept of Derivative

Differentiation is the study of rates of change. Geometrically, the derivative of a function at a point is the slope of the tangent line to the curve at that point.

ℹ️

Definition (First Principle)

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

$\text{Slope \text{ of } tangent at } (x_0, y_0) = f'(x_0) = \tan \theta$

Geometric Meaning of Derivative

Distance $h$ (Move slider to reduce $h$)

h = 0.1 (Tangent)h = 2.0 (Secant)

Current h:2.0

Slope of Secant PQ:4.000

Slope of Tangent (Actual):2.000

As $h \to 0$, Point Q moves closer to P, and the Secant slope approaches 2.0 (Tangent slope).

2. Standard Rules & Results

Algebraic Functions

\frac{d}{dx}(x^n)

→

nx^{n-1}

Ex:

\frac{d}{dx}(x^5) = 5x^4

\frac{d}{dx}((ax+b)^n)

→

n(ax+b)^{n-1} \cdot a

Ex:

\frac{d}{dx}((2x+3)^5) = 10(2x+3)^4

\frac{d}{dx}([f(x)]^n)

→

n[f(x)]^{n-1} f'(x)

Ex:

\frac{d}{dx}(\sin^3 x) = 3\sin^2 x \cos x

\frac{d}{dx}(a_n x^n + \dots)

→

na_n x^{n-1} + \dots

Ex:

\frac{d}{dx}(3x^2+5x+1) = 6x+5

\frac{d}{dx}(\text{constant})

→

0

Ex:

\frac{d}{dx}(100) = 0

\frac{d}{dx}(\sqrt{x})

→

\frac{1}{2\sqrt{x}}

Ex:

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

\frac{d}{dx}(\frac{1}{x})

→

-\frac{1}{x^2}

Ex:

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

Exponential

\frac{d}{dx}(e^x)

→

e^x

Ex:

\frac{d}{dx}(e^{2x}) = 2e^{2x}

\frac{d}{dx}(a^x)

→

a^x \log_e a

Ex:

\frac{d}{dx}(2^x) = 2^x \log 2

Logarithmic

\frac{d}{dx}(\log_e x)

→

\frac{1}{x}

Ex:

\text{Base must be } e

\frac{d}{dx}(\log_a x)

→

\frac{1}{x \log_e a}

Ex:

\frac{d}{dx}(\log_{10} x) = \frac{1}{x \log 10}

Algebra of Derivatives

Sum/Difference

\frac{d}{dx}[u \pm v] = u' \pm v'

Ex:

\frac{d}{dx}(x^2 + \sin x) = 2x + \cos x

Product Rule

\frac{d}{dx}(uv) = u'v + uv'

Ex:

\frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cos x

Quotient Rule

\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}

Ex:

\frac{d}{dx}(\frac{\sin x}{x}) = \frac{x \cos x - \sin x}{x^2}

Chain Rule

\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}

Ex:

\frac{d}{dx}(\log(\sin x)) = \cot x

3. Advanced Differentiation Techniques

Chain Rule

\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Used for composite functions. "Differentiate outside, then inside".

Example:

\frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x

Parametric Forms

\text{If } x = f(t), y = g(t), \text{ then } \frac{dy}{dx} = \frac{g'(t)}{f'(t)}

When x and y are both functions of a parameter 't'.

Example:

x=at^2, y=2at \implies \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}

Logarithmic Diff

\text{If } y = u^v, \log y = v \log u \implies \frac{1}{y} y' = v' \log u + v \frac{u'}{u}

Essential for functions of type $y = [f(x)]^{g(x)}$ .

Example:

y=x^x \implies y' = x^x(1+\log x)

4. Applications & Rate Measure

Rate Measure

Derivative represents instantaneous rate of change.

\frac{dy}{dt} = \text{Rate \text{ of } change \text{ of } } y \text{ w.r.t } t

💡 Useful Important Examples

Product + Chain Rule

Find derivative of $y = e^x \sin(x^2)$

y' = \frac{d}{dx}(e^x) \cdot \sin(x^2) + e^x \cdot \frac{d}{dx}(\sin(x^2))

= e^x \sin(x^2) + e^x \cos(x^2) \cdot 2x

= e^x(\sin(x^2) + 2x \cos(x^2))

Implicit Differentiation

Find $dy/dx$ if $x^3 + y^3 = 3axy$

3x^2 + 3y^2 y' = 3a(1 \cdot y + x y')

x^2 + y^2 y' = ay + ax y'

y'(y^2 - ax) = ay - x^2

y' = \frac{ay - x^2}{y^2 - ax}

Logarithmic Differentiation

Differentiate $y = (\sin x)^{\cos x}$

Take log:

\log y = \cos x \log(\sin x)

\implies\frac{1}{y}y' = -\sin x \log(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x

y' = (\sin x)^{\cos x}[\cos x \cot x - \sin x \log(\sin x)]

📝 Practice Zone

Concept Check Problems

Find the derivative of

f(x) = x^3 - 2x^2 + 5

x=1

✓ Show Solution

1. Differentiate:

f'(x) = 3x^2 - 4x

2. At

x=1

f'(1) = 3(1)^2 - 4(1) = 3 - 4 = -1

EASY

Differentiate

y = \log(\sin(x^2))

✓ Show Solution

Using Chain Rule:

y' = \frac{1}{\sin(x^2)} \cdot \frac{d}{dx}(\sin(x^2))

= \frac{1}{\sin(x^2)} \cdot \cos(x^2) \cdot \frac{d}{dx}(x^2)

= \cot(x^2) \cdot 2x

MEDIUM

Find

dy/dx

x^2 + y^2 = a^2

✓ Show Solution

Differentiate w.r.t x:

2x + 2y \frac{dy}{dx} = 0

2y \frac{dy}{dx} = -2x

\frac{dy}{dx} = -\frac{x}{y}

MEDIUM

Find

dy/dx

x = a\cos\theta, y = b\sin\theta

✓ Show Solution

dx/d\theta = -a\sin\theta

dy/d\theta = b\cos\theta

dy/dx = \frac{dy/d\theta}{dx/d\theta} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b}{a}\cot\theta

MEDIUM

🔥 CBSE Exam Priority Questions

Important Questions (Applied Maths)

y = x^x

, find

dy/dx

✓ Show Solution

Logarithmic Differentiation:

\log y = x \log x

2. Differentiate:

\frac{1}{y} y' = 1 \cdot \log x + x \cdot \frac{1}{x}

\frac{1}{y} y' = \log x + 1

y' = x^x(1 + \log x)

HARD

Find the rate of change of area of a circle w.r.t radius when

r = 5

cm.

✓ Show Solution

1. Area

A = \pi r^2

2. Rate w.r.t

r

\frac{dA}{dr} = 2\pi r

3. At

r=5

\frac{dA}{dr} = 10\pi

^2

/cm

EASY

y = \sqrt{\sin x + \sqrt{\sin x + ... \infty}}

, prove that

\frac{dy}{dx} = \frac{\cos x}{2y-1}

✓ Show Solution

1. Given

y = \sqrt{\sin x + y}

2. Square both sides:

y^2 = \sin x + y

3. Differentiate:

2y y' = \cos x + y'

y'(2y - 1) = \cos x \implies y' = \frac{\cos x}{2y-1}

HARD

Find the slope of tangent to the curve

y = 3x^4 - 4x

x=4

✓ Show Solution

y' = 12x^3 - 4

2. At

x=4

m = 12(4)^3 - 4 = 12(64) - 4 = 768 - 4 = 764

MEDIUM

x = a(t + \sin t), y = a(1 - \cos t)

, find

d^2y/dx^2

t=\pi/2

✓ Show Solution

dx/dt = a(1+\cos t), dy/dt = a\sin t

dy/dx = \frac{a\sin t}{a(1+\cos t)} = \tan(t/2)

\frac{d^2y}{dx^2} = \sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{dt}{dx}

4. At

t=\pi/2

dy/dx = 1

\frac{dt}{dx} = \frac{1}{a(1+0)} = 1/a

. Ans:

\frac{1}{2a}(\sqrt{2})^2 = 1/a

HARD

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1. Concept of Derivative

Differentiation is the study of rates of change. Geometrically, the derivative of a function at a point is the slope of the tangent line to the curve at that point.

ℹ️

Definition (First Principle)

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

$\text{Slope \text{ of } tangent at } (x_0, y_0) = f'(x_0) = \tan \theta$

Geometric Meaning of Derivative

Distance $h$ (Move slider to reduce $h$)

h = 0.1 (Tangent)h = 2.0 (Secant)

Current h:2.0

Slope of Secant PQ:4.000

Slope of Tangent (Actual):2.000

As $h \to 0$, Point Q moves closer to P, and the Secant slope approaches 2.0 (Tangent slope).

2. Standard Rules & Results

Algebraic Functions

\frac{d}{dx}(x^n)

→

nx^{n-1}

Ex:

\frac{d}{dx}(x^5) = 5x^4

\frac{d}{dx}((ax+b)^n)

→

n(ax+b)^{n-1} \cdot a

Ex:

\frac{d}{dx}((2x+3)^5) = 10(2x+3)^4

\frac{d}{dx}([f(x)]^n)

→

n[f(x)]^{n-1} f'(x)

Ex:

\frac{d}{dx}(\sin^3 x) = 3\sin^2 x \cos x

\frac{d}{dx}(a_n x^n + \dots)

→

na_n x^{n-1} + \dots

Ex:

\frac{d}{dx}(3x^2+5x+1) = 6x+5

\frac{d}{dx}(\text{constant})

→

0

Ex:

\frac{d}{dx}(100) = 0

\frac{d}{dx}(\sqrt{x})

→

\frac{1}{2\sqrt{x}}

Ex:

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

\frac{d}{dx}(\frac{1}{x})

→

-\frac{1}{x^2}

Ex:

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

Exponential

\frac{d}{dx}(e^x)

→

e^x

Ex:

\frac{d}{dx}(e^{2x}) = 2e^{2x}

\frac{d}{dx}(a^x)

→

a^x \log_e a

Ex:

\frac{d}{dx}(2^x) = 2^x \log 2

Logarithmic

\frac{d}{dx}(\log_e x)

→

\frac{1}{x}

Ex:

\text{Base must be } e

\frac{d}{dx}(\log_a x)

→

\frac{1}{x \log_e a}

Ex:

\frac{d}{dx}(\log_{10} x) = \frac{1}{x \log 10}

Algebra of Derivatives

Sum/Difference

\frac{d}{dx}[u \pm v] = u' \pm v'

Ex:

\frac{d}{dx}(x^2 + \sin x) = 2x + \cos x

Product Rule

\frac{d}{dx}(uv) = u'v + uv'

Ex:

\frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cos x

Quotient Rule

\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}

Ex:

\frac{d}{dx}(\frac{\sin x}{x}) = \frac{x \cos x - \sin x}{x^2}

Chain Rule

\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}

Ex:

\frac{d}{dx}(\log(\sin x)) = \cot x

3. Advanced Differentiation Techniques

Chain Rule

\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Used for composite functions. "Differentiate outside, then inside".

Example:

\frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x

Parametric Forms

\text{If } x = f(t), y = g(t), \text{ then } \frac{dy}{dx} = \frac{g'(t)}{f'(t)}

When x and y are both functions of a parameter 't'.

Example:

x=at^2, y=2at \implies \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}

Logarithmic Diff

\text{If } y = u^v, \log y = v \log u \implies \frac{1}{y} y' = v' \log u + v \frac{u'}{u}

Essential for functions of type $y = [f(x)]^{g(x)}$ .

Example:

y=x^x \implies y' = x^x(1+\log x)

4. Applications & Rate Measure

Rate Measure

Derivative represents instantaneous rate of change.

\frac{dy}{dt} = \text{Rate \text{ of } change \text{ of } } y \text{ w.r.t } t

💡 Useful Important Examples

Product + Chain Rule

Find derivative of $y = e^x \sin(x^2)$

y' = \frac{d}{dx}(e^x) \cdot \sin(x^2) + e^x \cdot \frac{d}{dx}(\sin(x^2))

= e^x \sin(x^2) + e^x \cos(x^2) \cdot 2x

= e^x(\sin(x^2) + 2x \cos(x^2))

Implicit Differentiation

Find $dy/dx$ if $x^3 + y^3 = 3axy$

3x^2 + 3y^2 y' = 3a(1 \cdot y + x y')

x^2 + y^2 y' = ay + ax y'

y'(y^2 - ax) = ay - x^2

y' = \frac{ay - x^2}{y^2 - ax}

Logarithmic Differentiation

Differentiate $y = (\sin x)^{\cos x}$

Take log:

\log y = \cos x \log(\sin x)

\implies\frac{1}{y}y' = -\sin x \log(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x

y' = (\sin x)^{\cos x}[\cos x \cot x - \sin x \log(\sin x)]

📝 Practice Zone

Concept Check Problems

Find the derivative of

f(x) = x^3 - 2x^2 + 5

x=1

✓ Show Solution

1. Differentiate:

f'(x) = 3x^2 - 4x

2. At

x=1

f'(1) = 3(1)^2 - 4(1) = 3 - 4 = -1

EASY

Differentiate

y = \log(\sin(x^2))

✓ Show Solution

Using Chain Rule:

y' = \frac{1}{\sin(x^2)} \cdot \frac{d}{dx}(\sin(x^2))

= \frac{1}{\sin(x^2)} \cdot \cos(x^2) \cdot \frac{d}{dx}(x^2)

= \cot(x^2) \cdot 2x

MEDIUM

Find

dy/dx

x^2 + y^2 = a^2

✓ Show Solution

Differentiate w.r.t x:

2x + 2y \frac{dy}{dx} = 0

2y \frac{dy}{dx} = -2x

\frac{dy}{dx} = -\frac{x}{y}

MEDIUM

Find

dy/dx

x = a\cos\theta, y = b\sin\theta

✓ Show Solution

dx/d\theta = -a\sin\theta

dy/d\theta = b\cos\theta

dy/dx = \frac{dy/d\theta}{dx/d\theta} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b}{a}\cot\theta

MEDIUM

🔥 CBSE Exam Priority Questions

Important Questions (Applied Maths)

y = x^x

, find

dy/dx

✓ Show Solution

Logarithmic Differentiation:

\log y = x \log x

2. Differentiate:

\frac{1}{y} y' = 1 \cdot \log x + x \cdot \frac{1}{x}

\frac{1}{y} y' = \log x + 1

y' = x^x(1 + \log x)

HARD

Find the rate of change of area of a circle w.r.t radius when

r = 5

cm.

✓ Show Solution

1. Area

A = \pi r^2

2. Rate w.r.t

r

\frac{dA}{dr} = 2\pi r

3. At

r=5

\frac{dA}{dr} = 10\pi

^2

/cm

EASY

y = \sqrt{\sin x + \sqrt{\sin x + ... \infty}}

, prove that

\frac{dy}{dx} = \frac{\cos x}{2y-1}

✓ Show Solution

1. Given

y = \sqrt{\sin x + y}

2. Square both sides:

y^2 = \sin x + y

3. Differentiate:

2y y' = \cos x + y'

y'(2y - 1) = \cos x \implies y' = \frac{\cos x}{2y-1}

HARD

Find the slope of tangent to the curve

y = 3x^4 - 4x

x=4

✓ Show Solution

y' = 12x^3 - 4

2. At

x=4

m = 12(4)^3 - 4 = 12(64) - 4 = 768 - 4 = 764

MEDIUM

x = a(t + \sin t), y = a(1 - \cos t)

, find

d^2y/dx^2

t=\pi/2

✓ Show Solution

dx/dt = a(1+\cos t), dy/dt = a\sin t

dy/dx = \frac{a\sin t}{a(1+\cos t)} = \tan(t/2)

\frac{d^2y}{dx^2} = \sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{dt}{dx}

4. At

t=\pi/2

dy/dx = 1

\frac{dt}{dx} = \frac{1}{a(1+0)} = 1/a

. Ans:

\frac{1}{2a}(\sqrt{2})^2 = 1/a

HARD