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Calculus - Derivatives of known functions – Rules

Grade 12IB_AA

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

🔑Concepts

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The power rule describes how the slope of a polynomial function y=xny = x^n changes. Geometrically, the derivative dydx=nxn−1\frac{dy}{dx} = nx^{n-1} represents the gradient of the tangent line at any point xx on the curve.

Graph of y = x^2 showing a tangent line at x=2 with gradient 4.
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The derivative of trigonometric functions like sin⁡x\sin x and cos⁡x\cos x are periodic. For example, ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x. This means the rate of change of the sine wave is exactly the value of the cosine wave at that same point.

Comparison of sine and cosine curves showing their derivative relationship.
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The Product Rule ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} is used when two functions are multiplied together. Think of it as the sum of one function times the rate of change of the other.

Flowchart representing the Product Rule process.
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The Quotient Rule ddx(uv)=vu′−uv′v2\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2} is applied to algebraic fractions. The order in the numerator is critical: start with the denominator function times the derivative of the numerator.

Graph of a rational function 1/x illustrating a case for the quotient rule.
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The Chain Rule dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} allows us to differentiate composite functions. It is often visualized as 'differentiating the outer layer and multiplying by the derivative of the inner layer'.

Flowchart of the composite function differentiation using the Chain Rule.

📐Formulae

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

ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}

ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x

ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x

ddx(tan⁡x)=1cos⁡2x\frac{d}{dx}(\tan x) = \frac{1}{\cos^2 x}

ddx(u⋅v)=udvdx+vdudx\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}

ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

💡Examples

Problem 1:

Differentiate f(x)=4x3−2ln⁡x+5exf(x) = 4x^3 - 2\ln x + 5e^x with respect to xx.

Solution:

f′(x)=12x2−2x+5exf'(x) = 12x^2 - \frac{2}{x} + 5e^x

Explanation:

Apply the power rule to 4x34x^3 to get 12x212x^2. Use the rule ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x} for the second term, and the rule that exe^x differentiates to itself for the third term.

Problem 2:

Find the derivative of y=x2sin⁡xy = x^2 \sin x.

Solution:

dydx=2xsin⁡x+x2cos⁡x\frac{dy}{dx} = 2x \sin x + x^2 \cos x

Explanation:

Use the Product Rule where u=x2u = x^2 and v=sin⁡xv = \sin x. Then u′=2xu' = 2x and v′=cos⁡xv' = \cos x. The result is u′v+uv′u'v + uv'.

Problem 3:

Differentiate y=(3x2+1)5y = (3x^2 + 1)^5 using the Chain Rule.

Solution:

dydx=5(3x2+1)4⋅(6x)=30x(3x2+1)4\frac{dy}{dx} = 5(3x^2 + 1)^4 \cdot (6x) = 30x(3x^2 + 1)^4

Explanation:

Let the inner function be u=3x2+1u = 3x^2 + 1. Then y=u5y = u^5. Differentiate the outer function dydu=5u4\frac{dy}{du} = 5u^4 and multiply by the derivative of the inner function dudx=6x\frac{du}{dx} = 6x.

Problem 4:

Find the gradient of the tangent to f(x)=cos⁡xxf(x) = \frac{\cos x}{x} at x=πx = \pi.

Solution:

f′(x)=−xsin⁡x−cos⁡xx2f'(x) = \frac{-x\sin x - \cos x}{x^2} At x=πx = \pi: f′(π)=−πsin⁡π−cos⁡ππ2=0−(−1)π2=1π2f'(\pi) = \frac{-\pi \sin \pi - \cos \pi}{\pi^2} = \frac{0 - (-1)}{\pi^2} = \frac{1}{\pi^2}

Explanation:

Apply the Quotient Rule where u=cos⁡xu = \cos x and v=xv = x. Then substitute x=πx = \pi into the resulting derivative function. Note that sin⁡π=0\sin \pi = 0 and cos⁡π=−1\cos \pi = -1.

Problem 5:

Find the derivative of the function f(x)=x2x+1f(x) = \frac{x^2}{x + 1} and determine the value of f′(1)f'(1).

Graph of f(x) = x^2 / (x+1) showing the tangent line at x=1.

Solution:

Let u=x2u = x^2 and v=x+1v = x + 1. Then dudx=2x\frac{du}{dx} = 2x and dvdx=1\frac{dv}{dx} = 1. Using the Quotient Rule: f′(x)=vdudx−udvdxv2f'(x) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} f′(x)=(x+1)(2x)−(x2)(1)(x+1)2f'(x) = \frac{(x+1)(2x) - (x^2)(1)}{(x+1)^2} f′(x)=2x2+2x−x2(x+1)2=x2+2x(x+1)2f'(x) = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2} Substitute x=1x = 1: f′(1)=12+2(1)(1+1)2=1+24=34f'(1) = \frac{1^2 + 2(1)}{(1+1)^2} = \frac{1+2}{4} = \frac{3}{4}

Explanation:

We identify the function as a quotient of two simpler functions. Applying the quotient rule formula and simplifying the numerator allows us to find the gradient function, which we then evaluate at the given point.

Problem 6:

Differentiate the composite function y=sin⁡(x2+3)y = \sin(x^2 + 3).

Graph of the composite function y = sin(x^2 + 3).

Solution:

This is a composite function of the form y=sin⁡(u)y = \sin(u) where u=x2+3u = x^2 + 3. Find the derivatives: dydu=cos⁡(u)\frac{dy}{du} = \cos(u) dudx=2x\frac{du}{dx} = 2x Using the Chain Rule: dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} dydx=cos⁡(x2+3)⋅(2x)\frac{dy}{dx} = \cos(x^2 + 3) \cdot (2x) dydx=2xcos⁡(x2+3)\frac{dy}{dx} = 2x \cos(x^2 + 3)

Explanation:

We use the Chain Rule to handle the 'function of a function'. First, we differentiate the outer sine function (which becomes cosine) and then multiply by the derivative of the inner polynomial expression.