Example 1: Equation of a tangent
Find the equation of the tangent to the curve y = x³ − 2x at the point where x = 1.
dy/dx = 3x² − 2- At
x = 1: slope= 3 − 2 = 1, andy = 1 − 2 = −1 y − (−1) = 1(x − 1)
Answer: y = x − 2
The derivative of f(x) is f′(x) = limh→0 [f(x + h) − f(x)] / h, the slope of the tangent to the curve. In M1 you need the idea and the notation (y′, f′(x), dy/dx); finding derivatives from first principles is not required.
f′(x) = limh→0 [f(x + h) − f(x)] / hx = x₀f′(x₀)y − f(x₀) = f′(x₀)(x − x₀)Find the equation of the tangent to the curve y = x³ − 2x at the point where x = 1.
dy/dx = 3x² − 2x = 1: slope = 3 − 2 = 1, and y = 1 − 2 = −1y − (−1) = 1(x − 1)Answer: y = x − 2
Find (a) limx→2 (x² − 4)/(x − 2) and (b) limx→∞ (3x² + 1)/(x² − 4).
(x² − 4)/(x − 2) = x + 2 for x ≠ 2, so the limit is 4x²: (3 + 1/x²)/(1 − 4/x²)x → ∞ the fractions tend to 0, so the limit is 3Answer: (a) 4; (b) 3
d/dx is an operator meaning "differentiate with respect to x". The curriculum notes single this out: dy/dx is one symbol, not dy divided by dx.x₀ into dy/dx for the slope and into the original equation for the y-coordinate. Using the derivative for both is a common slip.How to study: Click any question below to instantly load it on the coordinate grid and check your steps.