Differentiation is an important concept in calculus that helps us understand how a function changes. It is used to find the rate of change of one quantity with respect to another. For example, if we want to know how quickly the distance is changing with respect to time, we use differentiation.
The process of differentiation gives us a derivative, which represents the slope of a curve at a given point. In simple words, it tells us how steep a graph is or how quickly something increases or decreases. Differentiation is widely used in many real-life situations, such as mathematics, physics, economics, speed, development, and motion.
In this topic, we will explore the basic rules of differentiation formulas, common formulas, solved examples, and different types of questions that are often shown in the exam. This step-by-step guide will make it easy for you to understand and practise different questions effectively.
Table of Contents
Differentiation is the process of finding the rate of change of a function. In simple words, it tells us how fast one quantity changes when another quantity changes.
For example:
If distance changes with time, differentiation helps us find speed.
If speed changes with time, differentiation helps us find acceleration.
ddx(xn)=n.xn−1
ddx(sinx)=cosx
ddx(cosx)=−sinx
ddx(tanx)=sec2x
ddx(Inx)=1x
ddx(ex)=ex
Product Rule: ddx(u.v)=u′v+uv′
Quotient Rule: ddx(uv)=u′v−uv′v2
Chain Rule: ddx(f(g(x)))=f′(g(x)).g′(x)
Apply the power rule: d/dx(xⁿ) = n*xⁿ⁻¹
Differentiate 7x³: 7 * 3x² = 21x²
Differentiate -5x: -5
Constant 6 becomes 0
Final Answer: f'(x) = 21x² - 5
Use the standard derivative: d/dx(cos x) = -sin x
Final Answer: dy/dx = -sin(x)
Use chain rule: d/dx[ln(u)] = 1/u * du/dx
Let u = 3x + 1, then du/dx = 3
Apply the rule: 1/(3x + 1) * 3
Final Answer: dy/dx = 3 / (3x + 1)
Use the product rule: d/dx(u * v) = u'v + uv'
Let u = x², v = e^x
u' = 2x, v' = e^x
Apply rule: 2x * e^x + x² * e^x
Factor if needed: dy/dx = e^x(2x + x²)
Use quotient rule: d/dx[u/v] = (u'v - uv') / v²
Let u = x² + 1, v = x - 2
u' = 2x, v' = 1
Apply: [(2x)(x - 2) - (x² + 1)(1)] / (x - 2)²
Simplify numerator: 2x² - 4x - x² - 1 = x² - 4x - 1
Final Answer: dy/dx = (x² - 4x - 1) / (x - 2)²
Use chain rule: d/dx[(sin x)²] = 2 * sin(x) * d/dx[sin(x)]
d/dx[sin(x)] = cos(x)
Final Answer: dy/dx = 2 * sin(x) * cos(x)
Rewrite as y = 5x⁻²
Use power rule: d/dx(xⁿ) = n*xⁿ⁻¹
dy/dx = 5 * (-2)x⁻³ = -10/x³
Final Answer: dy/dx = -10 / x³
Use the standard formula: d/dx(arctan x) = 1 / (1 + x²)
Final Answer: dy/dx = 1 / (1 + x²)
Use chain rule: d/dx[e^(2x)] = e^(2x) * d/dx[2x]
d/dx[2x] = 2
Final Answer: dy/dx = 2e^(2x)
Use chain rule: d/dx[u⁴] = 4u³ * du/dx
Let u = x² + 4x + 5, then du/dx = 2x + 4
dy/dx = 4(x² + 4x + 5)³ * (2x + 4)
Final Answer: dy/dx = 4(2x + 4)(x² + 4x + 5)³
1. Differentiate: f(x) = x⁵
Use the Power Rule: d/dx(xⁿ) = n·xⁿ⁻¹
f'(x) = 5·x⁴
Answer: f'(x) = 5x⁴
2. Differentiate: f(x) = 3x² + 2x + 1
d/dx(3x²) = 6x
d/dx(2x) = 2
d/dx(1) = 0
Answer: f'(x) = 6x + 2
3. Differentiate: y = sin(x)
d/dx[sin(x)] = cos(x)
Answer: dy/dx = cos(x)
4. Differentiate: y = ln(x)
d/dx[ln(x)] = 1/x
Answer: dy/dx = 1/x
5. Differentiate: y = e^x
d/dx[e^x] = e^x
Answer: dy/dx = e^x
6. Differentiate: y = x³ · sin(x)
Use the product rule: (uv)' = u'v + uv'
u = x³ → u' = 3x²
v = sin(x) → v' = cos(x)
dy/dx = 3x²·sin(x) + x³·cos(x)
Answer: dy/dx = 3x²·sin(x) + x³·cos(x)
7. Differentiate: y = (2x² + 1) / (x + 3)
Use Quotient Rule: (u/v)' = (u'v - uv') / v²
u = 2x² + 1 → u' = 4x
v = x + 3 → v' = 1
dy/dx = [(4x)(x + 3) - (2x² + 1)(1)] / (x + 3)²
Answer: dy/dx = [(4x)(x + 3) - (2x² + 1)] / (x + 3)²
8. Differentiate: y = (3x + 4)²
Use the chain rule: d/dx[u²] = 2u·du/dx
u = 3x + 4 → du/dx = 3
dy/dx = 2(3x + 4)·3 = 6(3x + 4)
Answer: dy/dx = 6(3x + 4)
9. Differentiate: y = √x
Rewrite: y = x^(1/2)
Use the power rule: d/dx(xⁿ) = n·xⁿ⁻¹
dy/dx = (1/2)x^(-1/2) = 1 / (2√x)
Answer: dy/dx = 1 / (2√x)
10. Differentiate: y = tan(x)
d/dx[tan(x)] = sec²(x)
Answer: dy/dx = sec²(x)
Differentiate: f(x) = 7x³ - 5x + 6
Find dy/dx: y = cos(x)
Differentiate: y = ln(3x + 1)
Find the derivative: y = x² * e^x
Differentiate: y = (x² + 1) / (x - 2)
Find dy/dx: y = (sin(x))²
Differentiate: y = 5 / x²
Find the derivative: y = arctan(x)
Differentiate: y = e^(2x)
Find dy/dx: y = (x² + 4x + 5)⁴
Answer: A function's rate of change can be measured via differentiation. When we calculate acceleration (rate of change of velocity) or velocity (rate of change of position), for example, it helps us comprehend how one quantity changes in relation to another. In order to analyse dynamic systems, differentiation is a fundamental idea in physics, economics, and engineering.
Answer: There are several practical uses for differentiation. It aids in the computation of acceleration, velocity, and speed in physics. It can be applied in economics to either maximise profit or minimise expenses. Differentiation aids in the modeling of population growth rates in biology. It is also frequently used in engineering to examine structural stability and forces.
Answer: Simple methods like the power rule can be used to differentiate simple functions, such as polynomials (x² for example). Logarithmic, exponential, and trigonometric functions are examples of complex functions that may involve a combination of rules, such as the quotient, product, or chain rules. Divide complicated functions into smaller components, then apply the relevant rule to each one.
Answer: The derivative provides information on a function's change over time, which is essential for resolving practical issues. In the corporate world, for example, it can assist in figuring out the best prices for goods. It indicates the speed of an object at any given moment in physics. It can be used to compute material stress and strain in engineering.
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