A key idea in calculus that deals with a function's rate of change is differentiation. With numerous applications in physics, engineering, economics, and other fields, it is among the most significant topics in mathematics. By calculating the rate at which functions change about a variable, differentiation enables us to comprehend how functions act. We will go over the fundamentals of differentiation as well as its rules, methods, and applications in this self-study book. You will be ready to handle differentiating issues and successfully apply the ideas by the end.
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Differentiation is the process of finding the derivative of a function. The derivative indicates how a function changes at any point and is denoted by dy/dx or f’(x). In other words, it measures how much a function's output value changes as the input value changes.
In physics, differentiation helps calculate velocity and acceleration from position functions.
In economics, it is used to find marginal cost and marginal revenue.
In biology, differentiation helps in understanding population growth rates.
Engineers use it to determine rates of change in stress and strain in materials.
GPS and tracking systems use differentiation to calculate speed and direction.
d/dx (xⁿ) = n * xⁿ⁻¹
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (tan x) = sec² x
d/dx (ln x) = 1/x
d/dx (e^x) = e^x
Product Rule: d/dx (u * v) = u'v + uv'
Quotient Rule: d/dx (u / v) = (u'v - uv') / v²
Chain Rule: d/dx 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 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 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 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
A key idea in calculus, differentiation has wide-ranging applications in numerous domains. You can solve challenging problems in physics, engineering, economics, and other fields if you have a solid grasp of differentiation's principles and methods. You will become more confident when tackling real-world challenges if you practise and become familiar with differentiation rules.
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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.
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 modelling of population growth rates in biology. It is also frequently used in engineering to examine structural stability and forces.
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.
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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