Differentiating Powers

The pattern from $x^{2}$

First principles gave the derivative of $x^{2}$ as 2x. Look at where the numbers went: the power 2 came down in front as a multiplier, and the power on x dropped by one, from 2 to 1.

The power 2 of $x^{2}$ comes down in front, and the power drops to 1, giving 2x.

First principles on $x^{3}$

Run the same steps on $f\left(x\right) = x^{3}$. Multiply out $\left(x + h\right)^{3} = \left(x + h\right)\left(x^{2} + 2xh + h^{2}\right) = x^{3} + 3x^{2}h + 3xh^{2} + h^{3}$. Subtract $x^{3}$ and the $x^{3}$ cancels, so the rise is $3x^{2}h + 3xh^{2} + h^{3}$.

Divide through by h: the chord gradient is $3x^{2} + 3xh + h^{2}$. As h tends to 0, both 3xh and $h^{2}$ tend to 0, so the derivative of $x^{3}$ is $3x^{2}$.

Check at x = 2, where $3x^{2} = 12$. With h = 0.01, $\left(2.01\right)^{3} = 8.120601$, so the chord gradient is (8.120601 − 8) ÷ 0.01 = 12.0601, which is $3x^{2} + 3xh + h^{2} = 12 + 0.06 + 0.0001$.

Chords on $y = x^{3}$ from x = 1, with h = 0.6, 0.3 and 0.1. Their gradients are $3 + 3h + h^{2}$: 5.16, 3.99 and 3.31, closing in on the tangent, whose gradient is 3.

The power comes down

So $x^{3}$ gives $3x^{2}$. The same two moves as for $x^{2}$: the power 3 comes down in front, and the power on x drops by one, from 3 to 2.

The power 3 of $x^{3}$ comes down in front, and the power drops to 2, giving $3x^{2}$.

Any power

The rule is: to differentiate $x^{n}$, multiply by the power n, then take one off the power. The derivative of $x^{n}$ is $nx^{n-1}$. So $x^{4}$ gives $4x^{3}$, and $x^{5}$ gives $5x^{4}$.

The reason is in the expansion of $\left(x + h\right)^{n}$. Its first two terms are always $x^{n} + nx^{n-1}h$, and every term after them has $h^{2}$, $h^{3}$ or a higher power of h. Subtract $x^{n}$ and divide by h: what is left is $nx^{n-1}$ plus terms that each still contain h, and those tend to 0 with h. For n = 2 that is 2x + h, and for n = 3 it is $3x^{2} + 3xh + h^{2}$.

Check $x^{5}$ at x = 1, where the rule gives 5. With h = 0.001, $\left(1.001^{5} - 1\right) / 0.001$ is about 5.010.

For any power n, the n comes down in front and the power drops to n − 1.

A number in front, and sums

A number multiplying a power multiplies every rise, and so it multiplies the gradient too. So $kx^{n}$ differentiates to $k \times nx^{n-1}$: the old power multiplies the number in front. For $3x^{4}$ that is 3 × 4 = 12 in front and $x^{3}$ behind, $12x^{3}$.

Check $3x^{4}$ at x = 2, where $12x^{3} = 12 \times 8 = 96$. With h = 0.001, $2.001^{4}$ is about 16.032024, so the chord gradient is 3 × (16.032024 − 16) ÷ 0.001, about 96.07.

Two special cases follow from the rule. x is $x^{1}$, so it differentiates to $1 \times x^{0} = 1$: the line y = x has gradient 1. A constant such as 5 never changes, so its graph is level and its derivative is 0.

A sum is differentiated one term at a time. For $y = 2 + 20x - 5x^{2}$, the 2 gives 0, 20x gives 20, and $-5x^{2}$ gives −5 × 2x = −10x, so $\frac{dy}{dx} = 20 - 10x$.

The usual mistakes

Bringing the power down but not dropping it. $3x^{4}$ does not give $12x^{4}$: the power must also come down by one, to $x^{3}$.

Dropping the power but not multiplying. $3x^{4}$ does not give $3x^{3}$: the old power 4 multiplies the 3 in front, giving $12x^{3}$.

Differentiating a constant to itself. The 2 in $2 + 20x - 5x^{2}$ gives 0, because a constant does not change.

A ball thrown up

In the application below, the height of a ball is $y = 2 + 20x - 5x^{2}$ meters after x seconds. Differentiating term by term gives its velocity, differentiating again gives its acceleration, and the velocity is 0 at the top of the flight.