Differentiating Powers

Multiply by the power, then drop it by one.

The pattern from x²

First principles gave the derivative of x² 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.

x2
2xthe power comes down, then drops by one

The power 2 of x² comes down in front, and the power drops to 1, giving 2x.

First principles on x³

Run the same steps on f(x) = x³. Multiply out (x + h)³ = (x + h)(x² + 2xh + h²) = x³ + 3x²h + 3xh² + h³. Subtract x³ and the x³ cancels, so the rise is 3x²h + 3xh² + h³.

Divide through by h: the chord gradient is 3x² + 3xh + h². As h tends to 0, both 3xh and h² tend to 0, so the derivative of x³ is 3x².

Check at x = 2, where 3x² = 12. With h = 0.01, (2.01)³ = 8.120601, so the chord gradient is (8.120601 − 8) ÷ 0.01 = 12.0601, which is 3x² + 3xh + h² = 12 + 0.06 + 0.0001.

Chords on y = x³ from x = 1, with h = 0.6, 0.3 and 0.1. Their gradients are 3 + 3h + h²: 5.16, 3.99 and 3.31, closing in on the tangent, whose gradient is 3.

The power comes down

So x³ gives 3x². The same two moves as for x²: the power 3 comes down in front, and the power on x drops by one, from 3 to 2.

x3
the power comes down, then drops by one

The power 3 of x³ comes down in front, and the power drops to 2, giving 3x².

Any power

The rule is: to differentiate xⁿ, multiply by the power n, then take one off the power. The derivative of xⁿ is nxⁿ⁻¹. So x⁴ gives 4x³, and x⁵ gives 5x⁴.

The reason is in the expansion of (x + h)ⁿ. Its first two terms are always xⁿ + nxⁿ⁻¹h, and every term after them has h², h³ or a higher power of h. Subtract xⁿ and divide by h: what is left is nxⁿ⁻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 3x² + 3xh + h².

Check x⁵ at x = 1, where the rule gives 5. With h = 0.001, (1.001⁵ − 1) / 0.001 is about 5.010.

xn
same two moves for any power

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 kxⁿ differentiates to k × nxⁿ⁻¹: the old power multiplies the number in front. For 3x⁴ that is 3 × 4 = 12 in front and x³ behind, 12x³.

Check 3x⁴ at x = 2, where 12x³ = 12 × 8 = 96. With h = 0.001, 2.001⁴ 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 x¹, so it differentiates to 1 × x⁰ = 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 − 5x², the 2 gives 0, 20x gives 20, and −5x² gives −5 × 2x = −10x, so dy/dx = 20 − 10x.

The usual mistakes

Bringing the power down but not dropping it. 3x⁴ does not give 12x⁴: the power must also come down by one, to x³.

Dropping the power but not multiplying. 3x⁴ does not give 3x³: the old power 4 multiplies the 3 in front, giving 12x³.

Differentiating a constant to itself. The 2 in 2 + 20x − 5x² 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 − 5x² 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.

Worked example: A Ball Thrown Straight Up: Its Velocity and Acceleration from Its Height

Question A ball is thrown straight up. After x seconds its height above the ground is y = 2 + 20x − 5x2 meters. (a) Find the velocity of the ball after 1 second and after 3 seconds, and say what each one tells you. (b) Find the acceleration of the ball, and its height at the moment it stops rising.

  1. 1.Differentiate term by term. The constant 2 gives 0, 20x gives 20 and −5x2 gives −10x, so the velocity is v = dydx = 20 − 10x m/s.

    051015202501234seconds after the throw, xheight (m), yy = 2 + 20x − 5x2v = dy/dx = 20 − 10x
    051015202501234seconds after the throw, xheight (m), yy = 2 + 20x − 5x2v = dy/dx = 20 − 10x
    Differentiate term by term: the velocity is v = dydx = 20 − 10x m/s.

  2. 2.At x = 1, v = 20 − 10 = 10. At x = 3, v = 20 − 30 = −10.

    051015202501234seconds after the throw, xheight (m), y10 m/s−10 m/sx = 1: v = 20 − 10 = 10x = 3: v = 20 − 30 = −10
    051015202501234seconds after the throw, xheight (m), y10 m/s−10 m/sx = 1: v = 20 − 10 = 10x = 3: v = 20 − 30 = −10
    At x = 1, v = 10, and at x = 3, v = −10. They are the gradients of the two tangents.

  3. 3.(a) After 1 second the ball is rising at 10 m/s. After 3 seconds the velocity is −10 m/s: the minus sign means that the height is decreasing, so the ball is falling at 10 m/s.

    051015202501234seconds after the throw, xheight (m), y10 m/s−10 m/safter 1 s: rising at 10 m/safter 3 s: falling at 10 m/s
    051015202501234seconds after the throw, xheight (m), y10 m/s−10 m/safter 1 s: rising at 10 m/safter 3 s: falling at 10 m/s
    (a) After 1 second the ball is rising at 10 m/s. After 3 seconds the velocity is −10 m/s: the ball is falling at 10 m/s.

  4. 4.Differentiate the velocity: d2ydx2 = −10. The acceleration is −10 m/s2 at every moment of the flight, directed downward. It is the pull of gravity.

    051015202501234seconds after the throw, xheight (m), ya = d2y/dx2= −10 m/s2the same at every moment: gravity
    051015202501234seconds after the throw, xheight (m), ya = d2y/dx2= −10 m/s2the same at every moment: gravity
    Differentiate again: d2ydx2 = −10. The acceleration is −10 m/s2 at every moment, the pull of gravity.

  5. 5.The ball stops rising when v = 0: 20 − 10x = 0, so x = 2. Its height then is y = 2 + 20 × 2 − 5 × 22 = 2 + 40 − 20 = 22 m.

    051015202501234seconds after the throw, xheight (m), y(2, 22)v = 0: 20 − 10x = 0, so x = 2y = 2 + 40 − 20 = 22 m
    051015202501234seconds after the throw, xheight (m), y(2, 22)v = 0: 20 − 10x = 0, so x = 2y = 2 + 40 − 20 = 22 m
    The ball stops rising when v = 0, at x = 2. There the tangent is flat, and y = 2 + 40 − 20 = 22 m.

  6. 6.(b) The acceleration is −10 m/s2, and the ball stops rising at a height of 22 m. Check: the height is 17 m at x = 1 and again at x = 3, because the flight is symmetrical about x = 2.

    051015202501234seconds after the throw, xheight (m), y(2, 22)acceleration −10 m/s2, top at 22 mcheck: y = 17 at x = 1 and at x = 3
    051015202501234seconds after the throw, xheight (m), y(2, 22)acceleration −10 m/s2, top at 22 mcheck: y = 17 at x = 1 and at x = 3
    (b) The acceleration is −10 m/s2, and the ball stops rising at 22 m. The flight is symmetrical about x = 2.

Answer: (a) 10 m/s upward after 1 second, and −10 m/s after 3 seconds, so the ball is falling at 10 m/s; (b) −10 m/s2, and a height of 22 m

Common mistakes

  • Treating the velocity −10 m/s as a slip in the working. A negative velocity is correct: it says that the height is decreasing, so the ball is on its way down.
  • Differentiating the constant 2 to 2, or dropping the power in −5x2 and writing −5. A constant does not change, so its derivative is 0, and x2 becomes 2x, so −5x2 becomes −10x.

More introduction to calculus problems, worked step by step →

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