🔢 Tangent Tactics: Ace Differentiation & Gradients!

Differentiation is the engine that powers much of calculus. **Being able to find the gradient of any tangent quickly** lets you describe how quantities change instantaneously – a key skill tested in O Level papers and essential for science, engineering, and economics.

Core Idea: From Polynomial to Gradient

For a simple polynomial

y = ax^n

, the derivative

\displaystyle \frac{dy}{dx}

tells us the gradient function. **Apply the power rule**: bring the power

n

down, multiply by

a

, and reduce the power by 1, giving

\displaystyle \frac{dy}{dx} = anx^{n-1}

. This new expression lets us calculate the gradient at any

x

-value instantly.

Key point 1: $\frac{d}{dx}(x^n) = nx^{n-1}$ (Power Rule).
Key point 2: To find the gradient of a tangent at $x = k$ , substitute $x = k$ into $\displaystyle \frac{dy}{dx}$ .

Worked Example: Gradient at a Point

Problem: Given

y = 3x^3 - 5x^2 + 4x - 7

, find (i)

\displaystyle \frac{dy}{dx}

and (ii) the gradient of the curve at the point where

x = 2

Step 1: Differentiate term-by-term:
$\frac{dy}{dx} = 9x^2 - 10x + 4.$
Step 2: Evaluate at $x = 2$ :
$\text{Gradient} = 9(2)^2 - 10(2) + 4 = 36 - 20 + 4 = 20.$

Practice differentiating a variety of polynomials and plug in different

x

-values to solidify your skills. **Consistent practice ensures exam success**—so grab past papers and test yourself today!