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Algebra - Differentiation

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Differentiation is the mathematical process used to find the rate of change of a function. In geometry, the derivative dydx\frac{dy}{dx} represents the gradient (slope) of the tangent to a curve at any point (x,y)(x, y).

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For any term in the form axnax^n, the derivative is found by multiplying the coefficient by the power and then decreasing the power by 11. This is known as the Power Rule.

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The derivative of a constant term (a number without a variable, such as 55 or −12-12) is always 00, because a constant value does not change.

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The derivative of a linear term axax is simply the coefficient aa, because the gradient of a straight line y=axy = ax is constant.

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To differentiate expressions with variables in the denominator or under a square root, first rewrite them using indices: 1xn=x−n\frac{1}{x^n} = x^{-n} and x=x12\sqrt{x} = x^{\frac{1}{2}}.

📐Formulae

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}

ddx(axn)=anxn−1\frac{d}{dx}(ax^n) = anx^{n-1}

ddx(c)=0\frac{d}{dx}(c) = 0

ddx(ax)=a\frac{d}{dx}(ax) = a

ddx(u±v)=dudx±dvdx\frac{d}{dx}(u \pm v) = \frac{du}{dx} \pm \frac{dv}{dx}

💡Examples

Problem 1:

Differentiate the function y=5x4−3x2+7x−8y = 5x^4 - 3x^2 + 7x - 8 with respect to xx.

Solution:

dydx=5(4)x4−1−3(2)x2−1+7(1)x1−1−0\frac{dy}{dx} = 5(4)x^{4-1} - 3(2)x^{2-1} + 7(1)x^{1-1} - 0 dydx=20x3−6x+7\frac{dy}{dx} = 20x^3 - 6x + 7

Explanation:

Apply the Power Rule to each term individually. 5x45x^4 becomes 20x320x^3, −3x2-3x^2 becomes −6x-6x, 7x7x becomes 77, and the constant −8-8 becomes 00.

Problem 2:

Find the derivative of f(x)=4x2+xf(x) = \frac{4}{x^2} + \sqrt{x}.

Solution:

First, rewrite the expression using negative and fractional indices: f(x)=4x−2+x12f(x) = 4x^{-2} + x^{\frac{1}{2}} Now differentiate using the Power Rule: f′(x)=4(−2)x−3+12x12−1f'(x) = 4(-2)x^{-3} + \frac{1}{2}x^{\frac{1}{2} - 1} f′(x)=−8x−3+12x−12f'(x) = -8x^{-3} + \frac{1}{2}x^{-\frac{1}{2}} Simplify back to fraction form: f′(x)=−8x3+12xf'(x) = -\frac{8}{x^3} + \frac{1}{2\sqrt{x}}

Explanation:

Before differentiating, move x2x^2 to the numerator as x−2x^{-2} and convert the root to a power of 12\frac{1}{2}. Then apply the rule nxn−1nx^{n-1}.

Problem 3:

Find the gradient of the curve y=2x3−5xy = 2x^3 - 5x at the point where x=2x = 2.

Solution:

Step 1: Find the derivative dydx\frac{dy}{dx} to get the gradient function. dydx=6x2−5\frac{dy}{dx} = 6x^2 - 5 Step 2: Substitute x=2x = 2 into the gradient function. Gradient=6(2)2−5\text{Gradient} = 6(2)^2 - 5 Gradient=6(4)−5=24−5=19\text{Gradient} = 6(4) - 5 = 24 - 5 = 19

Explanation:

The gradient of a curve at a specific point is the value of its derivative at that point. We find the general derivative first, then plug in the given xx-coordinate.