Differentiating powers of x

By the end of this lesson

By the end you can answer "find $\dfrac{dy}{dx}$ " for any mix of powers of $x$ , roots and fractions, and find the second derivative $\dfrac{d^2y}{dx^2}$ or $f''(x)$ .

What they ask

Find $\dfrac{dy}{dx}$ , find $f'(x)$ , differentiate with respect to $x$ ("with respect to $x$ " means $x$ is the letter that changes), or find $\dfrac{d^2y}{dx^2}$ .

The function is a mix of powers, roots and fractions, like $y = 4x^3 - 6\sqrt{x} + \dfrac{2}{x^2}$ .

The method

Same three steps for every term:

Rewrite every term as a number times a power of $x$ , roots and fractions included. This is the step people skip.
Power down, one off: multiply by the power, then take 1 off the power.
Tidy up: a number on its own disappears, and you can turn negative powers back into fractions.

In symbols: $y = kx^n$ gives $\dfrac{dy}{dx} = knx^{n-1}$ .

$k$ is the number in front.
$n$ is the power. It can be any rational number: a whole number, a negative or a fraction.
$\dfrac{dy}{dx}$ (read "d y by d x") is the gradient function from the last lesson.

Why trust it? Last lesson the chords gave the gradient of $y = x^3$ at $x = 2$ as $12$ . The rule gives $3x^2 = 3(4) = 12$ . The same match happens for every power (see the gradient-function picture below).

Worked example: The easiest one: nothing to rewrite

Find $\dfrac{dy}{dx}$ when $y = 2x^3 - 5x + 4$ .

Have a go first?

Quick check

Write an expressionQuestion 1 · Recall

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