What is the derivative of x^n using the power rule?

The power rule states that d/dx of x^n equals n times x^(n-1) for any real n. Example: d/dx of x^5 = 5 x^4.

TOPIC 39 OF 45

Derivatives, Algebra of Derivatives and Exercises

🎓 Class 11 Mathematics CBSE Theory Ch 12 — Limits and Derivatives ⏱ ~30 min

🌐 Language: [gtranslate]

🧠 AI-Powered MCQ Assessment ▲

This MCQ module is based on: Derivatives, Algebra of Derivatives and Exercises

No. of Questions:

Difficulty Level:

📐 Maths Assessment ▲

This mathematics assessment will be based on: Derivatives, Algebra of Derivatives and Exercises
Targeting Class 11 level in Calculus, with Advanced difficulty.

Question Type:

Number of Questions:

Upload Additional Content (Optional):

Upload images, PDFs, or Word documents to include their content in assessment generation.

12.5 Derivatives — Definition

Let \(f\) be a real-valued function defined in an open interval containing a point \(a\). The derivative^? of \(f\) at \(a\) is defined as: \[f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h},\] provided this limit exists. If the limit exists, \(f\) is said to be differentiable at \(a\), and \(f'(a)\) is also denoted by \(\left.\dfrac{df}{dx}\right|_{x=a}\) or \(\left.\dfrac{d}{dx}f(x)\right|_{x=a}\).

Derivative Function

Replacing the fixed number \(a\) by a variable \(x\), we get the derivative function: \[f'(x)=\frac{d}{dx}f(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.\] This gives the derivative at every point where the limit exists.

Fig 12.4: The derivative \(f'(x)\) is the slope of the tangent at \((x,f(x))\), obtained as the limit of secant slopes.

Example 13

Find the derivative of \(f(x)=x^2\) at \(x=3\) from first principles.

\(f'(3)=\lim_{h\to 0}\dfrac{(3+h)^2-9}{h}=\lim_{h\to 0}\dfrac{9+6h+h^2-9}{h}=\lim_{h\to 0}(6+h)=6\).

Example 14

Find \(\dfrac{d}{dx}(x^n)\) for a positive integer \(n\).

\(\dfrac{d}{dx}x^n=\lim_{h\to 0}\dfrac{(x+h)^n-x^n}{h}\). Expand by Binomial: \((x+h)^n=x^n+nx^{n-1}h+\binom{n}{2}x^{n-2}h^2+\dots+h^n\). So \((x+h)^n-x^n=nx^{n-1}h+h^2\cdot[\text{polynomial in }x,h]\). Divide by \(h\): \(nx^{n-1}+h\cdot[\dots]\to n\,x^{n-1}\) as \(h\to 0\). \[\boxed{\ \frac{d}{dx}(x^n)=n\,x^{n-1}\ }\] (This formula is, in fact, valid for all real \(n\).)

Example 15

Find the derivative of \(f(x)=\sin x\) from first principles.

\(f'(x)=\lim_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h}\). Use sum-to-product: \(\sin(x+h)-\sin x=2\cos\!\left(x+\tfrac{h}{2}\right)\sin\!\left(\tfrac{h}{2}\right)\). Divide by \(h\): \(\cos(x+\tfrac{h}{2})\cdot\dfrac{\sin(h/2)}{h/2}\to \cos x\cdot 1=\cos x\). Hence \(\dfrac{d}{dx}\sin x=\cos x\).

Example 16

Find \(\dfrac{d}{dx}\cos x\) from first principles.

\(\cos(x+h)-\cos x=-2\sin\!\left(x+\tfrac{h}{2}\right)\sin\!\left(\tfrac{h}{2}\right)\). Divide by \(h\): \(-\sin(x+\tfrac{h}{2})\cdot\dfrac{\sin(h/2)}{h/2}\to -\sin x\). So \(\dfrac{d}{dx}\cos x=-\sin x\).

12.6 Algebra of Derivatives

Differentiation Rules

Let \(u=f(x)\), \(v=g(x)\) be differentiable functions. Then:

Sum / Difference: \(\dfrac{d}{dx}(u\pm v)=\dfrac{du}{dx}\pm\dfrac{dv}{dx}\)
Product Rule: \(\dfrac{d}{dx}(uv)=\dfrac{du}{dx}\cdot v+u\cdot\dfrac{dv}{dx}\)
Quotient Rule: \(\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right)=\dfrac{\tfrac{du}{dx}\,v-u\,\tfrac{dv}{dx}}{v^2}\) for \(v\ne 0\)
Constant multiple: \(\dfrac{d}{dx}(k u)=k\,\dfrac{du}{dx}\)
Constant: \(\dfrac{d}{dx}(c)=0\)

Derivation of the Product Rule

Let \(F(x)=u(x)v(x)\). Then \(F(x+h)-F(x)=u(x+h)v(x+h)-u(x)v(x)\) \(=u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)\) \(=u(x+h)[v(x+h)-v(x)]+v(x)[u(x+h)-u(x)]\). Divide by \(h\) and take the limit. Using continuity of \(u\) (differentiability ⟹ continuity): \[F'(x)=u(x)\,v'(x)+v(x)\,u'(x).\]

Example 17

Differentiate \(y=(x^2+1)(x^3+3)\).

Product rule: \(y'=(2x)(x^3+3)+(x^2+1)(3x^2)=2x^4+6x+3x^4+3x^2=5x^4+3x^2+6x\).

Example 18

Differentiate \(y=\dfrac{x+1}{x-1}\).

Quotient rule with \(u=x+1, v=x-1\): \(y'=\dfrac{(1)(x-1)-(x+1)(1)}{(x-1)^2}=\dfrac{x-1-x-1}{(x-1)^2}=\dfrac{-2}{(x-1)^2}\).

Example 19

Differentiate \(y=\tan x\).

\(\tan x=\dfrac{\sin x}{\cos x}\). Quotient rule: \(y'=\dfrac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x}=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}=\dfrac{1}{\cos^2 x}=\sec^2 x\).

Standard Derivatives

f(x)	\(x^n\)	\(\sin x\)	\(\cos x\)	\(\tan x\)	c (constant)	\(ax+b\)
f '(x)	\(n x^{n-1}\)	\(\cos x\)	\(-\sin x\)	\(\sec^2 x\)	0	a

Exercise 12.2 (Selected Problems)

Q1. Find the derivative of \(x^2-2\) at \(x=10\).

\(f'(x)=2x\); \(f'(10)=20\).

Q2. Find the derivative of \(99x\) at \(x=100\).

\(f'(x)=99\) everywhere, so \(f'(100)=99\).

Q3. Find \(\dfrac{d}{dx}(x^3-27)\) at \(x=1\).

\(3x^2\) at \(x=1\) = 3.

Q4. Differentiate \((x-a)(x-b)\) with respect to \(x\).

Product rule: \((1)(x-b)+(x-a)(1)=2x-a-b\).

Q5. Differentiate \(y=(5x^3+3x-1)(x-1)\).

\(y'=(15x^2+3)(x-1)+(5x^3+3x-1)(1)=15x^3-15x^2+3x-3+5x^3+3x-1=20x^3-15x^2+6x-4\).

Q6. Differentiate \(y=x^{-3}(5+3x)\).

Expand: \(y=5x^{-3}+3x^{-2}\). \(y'=-15x^{-4}-6x^{-3}=-\dfrac{15}{x^4}-\dfrac{6}{x^3}\).

Q7. Find \(\dfrac{d}{dx}\!\left(\dfrac{1}{x}\right)\).

\(x^{-1}\); derivative = \(-x^{-2}=-\dfrac{1}{x^2}\).

Q8. Differentiate \(y=\sin x+\cos x\).

\(y'=\cos x-\sin x\).

Q9. Differentiate \(y=x\sin x\).

Product rule: \(y'=(1)\sin x+x\cos x=\sin x+x\cos x\).

Q10. Differentiate \(y=\dfrac{\sin x}{x}\).

Quotient rule: \(y'=\dfrac{x\cos x-\sin x}{x^2}\).

Q11. For \(f(x)=\dfrac{ax+b}{cx+d}\) (\(c\ne 0\), \(cx+d\ne 0\)), find \(f'(x)\).

Quotient rule: \(f'(x)=\dfrac{a(cx+d)-(ax+b)c}{(cx+d)^2}=\dfrac{ad-bc}{(cx+d)^2}\).

Activity: First Principles on a Spreadsheet

L3 Apply

Materials: Spreadsheet (or calculator), paper.

Predict: For \(f(x)=x^3\), compute \(\dfrac{f(2+h)-f(2)}{h}\) for \(h=0.1, 0.01, 0.001\). What value are you approaching?

Create columns: \(h\), \(f(2+h)\), \(f(2)\), \(\dfrac{f(2+h)-f(2)}{h}\).
Use \(f(x)=x^3\) and decrease \(h\) progressively.
Record the successive quotient values.
Compare with \(f'(2)\) from the formula \(f'(x)=3x^2\).
Repeat the activity with \(f(x)=\sin x\) at \(x=\pi/3\) and \(h=0.01\).

For \(x^3\) at \(x=2\), quotients approach 12, matching \(f'(2)=3\cdot 4=12\). For \(\sin x\) at \(x=\pi/3\), quotients approach \(\cos(\pi/3)=0.5\). This confirms that calculating a derivative from first principles and from the derivative formula give the same answer.

Competency-Based Questions

Scenario: A cyclist's position is \(s(t)=t^3-6t^2+9t\) (metres) after \(t\) seconds. We investigate her velocity \(s'(t)\) and acceleration \(s''(t)\).

Q1. Her velocity at \(t=2\) s is:

L3 Apply

(a) −3 m/s
(b) 0 m/s
(c) 3 m/s
(d) 9 m/s

Answer: (a) −3 m/s. \(s'(t)=3t^2-12t+9\); \(s'(2)=12-24+9=-3\). Negative sign indicates motion in the negative direction.

Q2. Analyse: at what moments is she momentarily at rest? What happens to her position at those instants?

L4 Analyse

Answer: Set \(s'(t)=0\): \(3t^2-12t+9=0 \Rightarrow t^2-4t+3=0 \Rightarrow (t-1)(t-3)=0\). So \(t=1\) and \(t=3\). At \(t=1\): \(s=1-6+9=4\) m. At \(t=3\): \(s=27-54+27=0\) m. These are turning points: the cyclist reverses direction at each.

Q3. Evaluate the claim: "The cyclist accelerates uniformly throughout." Use \(s''(t)\) to support or refute.

L5 Evaluate

Answer: \(s''(t)=\dfrac{d}{dt}(3t^2-12t+9)=6t-12\). This is not constant, hence the acceleration changes with time — the claim is false. Acceleration is negative for \(t<2\), zero at \(t=2\), positive for \(t>2\).

Q4. Create a different position function \(s(t)\) (a polynomial of degree 3) for which the cyclist comes to rest exactly at \(t=2\) and \(t=5\). Verify your answer by differentiation.

L6 Create

Sample: Need \(s'(t)=k(t-2)(t-5)=k(t^2-7t+10)\). Integrate (reverse): \(s(t)=k\left(\tfrac{t^3}{3}-\tfrac{7t^2}{2}+10t\right)+C\). Pick \(k=6, C=0\): \(s(t)=2t^3-21t^2+60t\). Check: \(s'(t)=6t^2-42t+60=6(t-2)(t-5)\) — zero at \(t=2,5\). ✓

Assertion–Reason Questions

Assertion (A): \(\dfrac{d}{dx}\tan x=\sec^2 x\).
Reason (R): \(\tan x=\dfrac{\sin x}{\cos x}\), and the quotient rule applies.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a). Quotient rule yields \((\cos^2x+\sin^2x)/\cos^2x=\sec^2x\). R explains A.

Assertion (A): The derivative of a sum is the sum of derivatives.
Reason (R): The limit of a sum equals the sum of limits.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a). Since the derivative is defined as a limit of a difference quotient, the sum law for limits transfers directly to the sum law for derivatives.

Assertion (A): \(\dfrac{d}{dx}(uv)=u'v'\).
Reason (R): Differentiation distributes over multiplication just as over addition.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d?) Both A and R are false. Correct product rule: \((uv)'=u'v+uv'\), not \(u'v'\). Differentiation does NOT distribute over multiplication. Best choice among given options: neither A nor R is true → marked as (d)-style rejection; the intended key is that both are false. Choose "none" if available.

Frequently Asked Questions

What is the derivative from first principles?

f '(x) is defined as the limit as h approaches 0 of [ f(x + h) - f(x) ] / h. This is also called the derivative by ab initio or by definition.

What is the product rule for derivatives?

If u and v are differentiable, (u v)' = u' v + u v'. The derivative of a product is not the product of the derivatives.

What is the quotient rule?

(u/v)' = (u' v - u v') / v squared, provided v is not 0. This gives the derivative of a ratio of two functions.

What is the derivative of a constant?

The derivative of any constant c is 0. This follows because a constant function has zero rate of change.

What is the summary of Class 11 Chapter 12 Limits and Derivatives?

Chapter 12 introduces limits, algebra of limits, standard trig limits, the intuitive idea of a derivative, first-principles definition, the power rule and the algebra of derivatives.

How important is Chapter 12 for Class 12 and JEE?

Extremely important. It is the foundation for Class 12 Calculus (continuity, differentiation, integration) and directly contributes 8-12 marks in JEE Main every year.

AI Tutor

Mathematics Class 11

Ready

Hi! 👋 I'm Gaura, your AI Tutor for Derivatives, Algebra of Derivatives and Exercises. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.