A-level Mathematics

7.3 Differentiating powers, exponentials, trigonometric functions and logarithms — Learn

Prerequisites: the laws of indices for rational exponents; the derivative as a limit from Section 7.2; the graphs of exe^x and ln⁡x\ln x from Chapter 6; the trigonometric functions from Chapter 5.

By the end of this section you can: differentiate xnx^n for any rational nn; differentiate a constant multiple, a sum and a difference of such terms; differentiate ekxe^{kx}, akxa^{kx}, sin⁡kx\sin kx, cos⁡kx\cos kx, tan⁡kx\tan kx and ln⁡x\ln x; and choose the right rule for a given expression.

1. The power rule

For any rational number nn, the derivative of xnx^n is

ddx(xn)=nxn−1.\frac{d}{dx}\left(x^n\right) = n x^{n-1}.

Multiply by the old power, then reduce the power by one. The rule holds for whole numbers, fractions and negative numbers.

Example 1. Differentiate y=x4y = x^4.

Multiply by 44 and reduce the power: dydx=4x3\dfrac{dy}{dx} = 4x^3.

Example 2. Differentiate y=xy = \sqrt{x}.

Write the root as a power: y=x1/2y = x^{1/2}. Then

dydx=12x−1/2=12x.\frac{dy}{dx} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.

Example 3. Differentiate y=1xy = \dfrac{1}{x}.

Write the fraction as a power: y=x−1y = x^{-1}. Then

dydx=−1x−2=−1x2.\frac{dy}{dx} = -1x^{-2} = -\frac{1}{x^2}.

The derivative is negative everywhere, so the curve falls on each side of the break at x=0x = 0.

A constant multiple, a sum and a difference are differentiated term by term:

ddx(axn+bxm)=anxn−1+bmxm−1.\frac{d}{dx}\left(ax^n + bx^m\right) = anx^{n-1} + bmx^{m-1}.

Example 4. Differentiate y=4x3−6x+1xy = 4x^3 - 6x + \dfrac{1}{x}.

Rewrite 1x\dfrac{1}{x} as x−1x^{-1} and differentiate each term:

dydx=12x2−6−x−2=12x2−6−1x2.\frac{dy}{dx} = 12x^2 - 6 - x^{-2} = 12x^2 - 6 - \frac{1}{x^2}.

Example 5. Differentiate y=2x3/2+3x2y = 2x^{3/2} + \dfrac{3}{x^2}.

Rewrite 3x2\dfrac{3}{x^2} as 3x−23x^{-2}. Then

dydx=2×32x1/2−6x−3=3x1/2−6x3=3x−6x3.\frac{dy}{dx} = 2 \times \tfrac{3}{2}x^{1/2} - 6x^{-3} = 3x^{1/2} - \frac{6}{x^3} = 3\sqrt{x} - \frac{6}{x^3}.

2. Exponential functions

For a constant kk,

ddx(ekx)=kekx.\frac{d}{dx}\left(e^{kx}\right) = k e^{kx}.

The exponential function is special because its derivative has the same shape as the function itself. The constant kk appears in front. When k=1k = 1 the derivative of exe^x is exe^x.

For a positive base aa,

ddx(akx)=kakxln⁡a.\frac{d}{dx}\left(a^{kx}\right) = k a^{kx}\ln a.

The factor ln⁡a\ln a appears because akx=ekxln⁡aa^{kx} = e^{kx\ln a}.

Example 6. Differentiate y=5e3x−2e−xy = 5e^{3x} - 2e^{-x}.

dydx=5×3e3x−2×(−1)e−x=15e3x+2e−x.\frac{dy}{dx} = 5 \times 3e^{3x} - 2 \times (-1)e^{-x} = 15e^{3x} + 2e^{-x}.

Example 7. Differentiate y=2xy = 2^x.

Here a=2a = 2 and k=1k = 1, so

dydx=2xln⁡2.\frac{dy}{dx} = 2^x \ln 2.

3. Trigonometric functions

For a constant kk,

ddx(sin⁡kx)=kcos⁡kx,ddx(cos⁡kx)=−ksin⁡kx.\frac{d}{dx}\left(\sin kx\right) = k\cos kx, \qquad \frac{d}{dx}\left(\cos kx\right) = -k\sin kx.

The derivative of sine is cosine. The derivative of cosine is minus sine, so the minus sign often goes missing.

The tangent has its own result:

ddx(tan⁡kx)=kcos⁡2kx=ksec⁡2kx.\frac{d}{dx}\left(\tan kx\right) = \frac{k}{\cos^2 kx} = k\sec^2 kx.

Example 8. Differentiate y=3sin⁡2x−4cos⁡xy = 3\sin 2x - 4\cos x.

dydx=3×2cos⁡2x−4×(−sin⁡x)=6cos⁡2x+4sin⁡x.\frac{dy}{dx} = 3 \times 2\cos 2x - 4 \times (-\sin x) = 6\cos 2x + 4\sin x.

Example 9. Differentiate y=tan⁡3xy = \tan 3x.

dydx=3sec⁡23x=3cos⁡23x.\frac{dy}{dx} = 3\sec^2 3x = \frac{3}{\cos^2 3x}.

4. The natural logarithm

For x>0x > 0,

ddx(ln⁡x)=1x.\frac{d}{dx}\left(\ln x\right) = \frac{1}{x}.

The rule also works for ln⁡kx\ln kx when k>0k > 0, because ln⁡kx=ln⁡k+ln⁡x\ln kx = \ln k + \ln x and ln⁡k\ln k is a constant. So

ddx(ln⁡5x)=1x.\frac{d}{dx}\left(\ln 5x\right) = \frac{1}{x}.

Example 10. Differentiate y=4ln⁡x−ln⁡3xy = 4\ln x - \ln 3x.

The first term gives 4x\dfrac{4}{x}, and the second gives 1x\dfrac{1}{x}:

dydx=4x−1x=3x.\frac{dy}{dx} = \frac{4}{x} - \frac{1}{x} = \frac{3}{x}.

5. Try it: the power rule explorer

Predict first. The figure shows y=xny = x^n with a slider for nn and a point you can move. Before you touch the controls, predict the gradient at x=1x = 1 for n=3n = 3, and predict whether the gradient at x=1x = 1 is larger or smaller when n=5n = 5. Write both predictions down.

Test. Set n=3n = 3 and move the point to x=1x = 1. Read the gradient. Change nn to 55 and read it again. Then set n=0.5n = 0.5 and check the gradient at x=1x = 1 and at x=4x = 4.

Explain. Write two sentences. The first explains why the gradient at x=1x = 1 equals nn for every curve y=xny = x^n. The second explains why a curve with a negative power nn has a negative gradient.

Watch out. These mistakes are common.

Increasing the power instead of reducing it. The power rule is xn→nxn−1x^n \to n x^{n-1}. The power goes down by one. Writing nxn+1n x^{n+1} is a different function.

Forgetting the constant in front of an exponential. The derivative of e3xe^{3x} is 3e3x3e^{3x}, not e3xe^{3x}. The chain rule, met formally in Section 7.7, is what produces the 33.

Dropping the minus sign on cosine. The derivative of cos⁡kx\cos kx is −ksin⁡kx-k\sin kx. Check the sign with k=1k = 1 at x=0x = 0: the curve y=cos⁡xy = \cos x is falling there, so its gradient must be negative.

Treating ln⁡x\ln x as a power. The derivative of ln⁡x\ln x is 1x\dfrac{1}{x}, which is not x0x^0 or 00. Re-check: differentiate xx, x0x^0 and ln⁡x\ln x separately and confirm that only the first gives 11.

Progress check

Return to the box in the chapter problem. Its volume is V=x(16−2x)(10−2x)V = x(16 - 2x)(10 - 2x).

  1. Expand the brackets to write VV as a cubic in xx.
  2. Differentiate VV with the power rule.
  3. Setting dVdx=0\dfrac{dV}{dx} = 0 gives a quadratic. Write that quadratic down.

The next sections find the value of xx and prove that it gives the greatest volume.

You should now be able to

Settings

Colour scheme

Applies to every book you read here.

Reading theme

Applies to every book you read here.