Differentiation explained: gradients and the power rule

Multiply by the power, then take one off the power.

Differentiate term by term; constants disappear.

Substitute the x-value into the derivative.

Example: for $y = x^2$, $\dfrac{dy}{dx} = 2x$. At $x = 3$, the gradient is $6$: the curve is rising 6 units for each 1 across, at that exact point.

The equation of the tangent there uses that gradient and the point $(3, 9)$: $y = 6x - 9$.

Where the gradient is zero, the curve is momentarily flat: a maximum or minimum (a turning point).

Example: $y = x^3 - 3x$. $\dfrac{dy}{dx} = 3x^2 - 3 = 0$ gives $x = 1$ or $x = -1$. The points are $(1, -2)$ and $(-1, 2)$.

To tell which is which, use the second derivative, $\dfrac{d^2y}{dx^2} = 6x$. At $x = 1$ it's positive (a minimum); at $x = -1$ it's negative (a maximum).

The derivative is a rate of change. If $s$ is distance and $t$ is time, $\dfrac{ds}{dt}$ is velocity; differentiate again for acceleration. In economics, the derivative of total cost is the marginal cost. Any time a question asks "how fast is it changing?", differentiation is the tool.

Rewrite everything as powers of $x$ before differentiating:

• $\sqrt{x} = x^{1/2}$, so its derivative is $\tfrac{1}{2}x^{-1/2}$.
• $\dfrac{1}{x^2} = x^{-2}$, so its derivative is $-2x^{-3}$.
• Expand brackets: $(x + 1)^2 = x^2 + 2x + 1$, so the derivative is $2x + 2$.

Our expanding brackets calculator helps with the last step.

• Forgetting to multiply by the power.
• Differentiating a constant to 1 instead of 0.
• Differentiating before rewriting roots and fractions as powers.
• Substituting into the original function instead of the derivative to find a gradient.

For the straight-line version of gradient, see our gradient calculator.

Integration undoes differentiation. To integrate $x^n$, add one to the power and divide by the new power: $\int x^n \, dx = \dfrac{x^{n+1}}{n+1} + c$. The $+ c$ appears because constants vanish when you differentiate. Integration finds areas under curves, just as differentiation finds gradients.

1. Differentiate $4x^3$.
2. Find the stationary point of $y = x^2 - 6x + 5$ and say whether it's a maximum or minimum.
3. Find the gradient of $y = x^3$ at $x = 2$.
4. Differentiate $6\sqrt{x}$.

Answers: 1. $12x^2$. 2. $\dfrac{dy}{dx} = 2x - 6 = 0$ gives $x = 3$, $y = -4$: a minimum (the $x^2$ term is positive). 3. $3x^2 = 12$. 4. $6x^{1/2}$ differentiates to $3x^{-1/2}$, or $\dfrac{3}{\sqrt{x}}$.

• Rewrite every term as a power of $x$ before differentiating.
• When asked for the equation of a tangent, find the gradient from $\tfrac{dy}{dx}$ and the point from the original equation.
• Always say whether a stationary point is a maximum or minimum, and how you know.