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Last updated on September 25, 2025
In calculus, differentiation and integration are two fundamental concepts. Differentiation helps find the rate of change of a function, while integration is used to find the area under curves. In this topic, we will learn the formulas for differentiation and integration.
Differentiation and integration are core concepts in calculus. Let’s learn the formulas to calculate derivatives and integrals.
Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change.
Basic differentiation formula: If \(y = f(x)\), then the derivative is \(\frac{dy}{dx} = f'(x)\)
Power rule: (frac{d}{dx}x^n = nx^{n-1})
Sum rule: (frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)) Product rule: (frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x))
Quotient rule: (frac{d}{dx}left[frac{f(x)}{g(x)}right] = frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2})
Integration is the process of finding the integral of a function, which represents the accumulated area under the curve.
Basic integration formula: If (F(x)) is an antiderivative of (f(x)), then (int f(x) dx = F(x) + C)
Power rule: (int x^n dx = frac{x^{n+1}}{n+1} + C)
Sum rule: (int [f(x) + g(x)] dx = int f(x) dx + int g(x) dx) Integration by parts: (int u dv = uv - int v du)
Substitution rule: If (u = g(x)), then (int f(g(x))g'(x) dx = int f(u) du)
In math and real life, we use differentiation and integration formulas to solve various problems involving rates of change and areas.
Here are some important points about differentiation and integration.
Students find calculus formulas tricky and confusing.
Here are some tips and tricks to master them.
Differentiation and integration play a major role in various real-life applications.
Here are some examples.
Students make errors when applying differentiation and integration formulas. Here are some mistakes and ways to avoid them.
Differentiate \(f(x) = 3x^4 - 2x^2 + x\)
The derivative is \(f'(x) = 12x^3 - 4x + 1\)
Using the power rule, differentiate each term: (f'(x) = 3 cdot 4x^{4-1} - 2 cdot 2x^{2-1} + 1x^{1-1})
Simplify to get \(f'(x) = 12x^3 - 4x + 1\)
Integrate \(\int (2x^3 - 5x + 4) dx\)
The integral is \(\frac{1}{2}x^4 - \frac{5}{2}x^2 + 4x + C\)
Using the power rule for integration: (int 2x3 dx = frac{2}{4}x4 = frac{1}{2}x4) (\int -5x dx = -frac{5}{2}x2) (int 4 dx = 4x\)
Combine and add the constant \(C\): (frac{1}{2}x4 - frac{5}{2}x2 + 4x + C)
Find the derivative of \(g(x) = x^2 \sin(x)\)
The derivative is \(g'(x) = 2x sin(x) + x2 \cos(x)\)
Apply the product rule: \(\frac{d}{dx}[x2 \sin(x)] = \frac{d}{dx}[x^2] \cdot \sin(x) + x2 cdot \frac{d}{dx}[\sin(x)]\)
Calculate each term: \(2x \sin(x) + x^2 \cos(x)\)
Integrate \(\int e^x \cos(x) dx\) using integration by parts
The solution involves multiple applications of integration by parts, resulting in a complex expression.
Integration by parts: (int u dv = uv - int v du)
Select (u = ex) and (dv = cos(x) dx), then differentiate and integrate: (du = ex dx), (v = sin(x))
Apply the formula: (int ex cos(x) dx = ex sin(x) - int sin(x) ex dx)
Repeat the process for the new integral.
Differentiate \(h(x) = \frac{2x + 3}{x^2}\)
The derivative is (h'(x) = frac{-2x2 - 4x - 3}{x4})
Use the quotient rule: (frac{d}{dx}left[frac{2x+3}{x2}right] = frac{(2)(x2) - (2x+3)(2x)}{(x^2)2})
Simplify to get \(\frac{-2x^2 - 4x - 3}{x^4}\)
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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