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Differentiating trig, exponentials and logs

The power rule ran Year 12; now the rest of the function shelf joins the calculus. Sine's gradient turns out to be cosine, e keeps its old promise, the logarithm's slope is a reciprocal, and every one of these facts is radian-powered machinery, not coincidence.

Year 12-13EDEXCEL 9MA0 7.1, 7.2

Builds on Differentiating powers of x and Radians, arcs and small angles.

IN THIS TOPIC

  • Differentiate sin kx, cos kx, tan kx, ekx, akx and ln x fluently.
  • Differentiate sin x from first principles using the small-angle approximations.
  • Find gradients and tangents on trig, exponential and log curves.

WHAT YOU PROBABLY THINK

The derivative of sin x is cos x, whatever unit the angle is in.

The full derivative shelf

Four families join the power rule, all on the must-learn list,

sin kx → kcos kx, cos kx → −ksin kxNOT IN THE BOOKLET — LEARN IT
ekx → kekx, ln x → 1xNOT IN THE BOOKLET — LEARN IT

with tan kx → k sec2 kx available from the booklet, and one more the specification names explicitly: akx differentiates to kakx ln a, the ln a being the price any base other than e pays.

The sine curve drawn above its own gradient function, which is the cosine curve: sine's hilltops and valleys sit exactly over cosine's zero crossings, and sine's steepest climbs over cosine's peaksy = sin xgradient: cos xflat points above, zero crossings below, in radians
FIG. 1Sine's slope readout is cosine: flat points of sin x sit exactly over the zeros of cos x. In radians, and only in radians.

WORKED EXAMPLE

A mixed bag, termwise

Differentiate y = 4 sin 3x − 2 cos 5x + e3x.

Termwise with the shelf: 4 sin 3x gives 12 cos 3x, and −2 cos 5x gives +10 sin 5x, the minus signs cancelling.

The exponential gives 3e3x.

dy/dx = 12 cos 3x + 10 sin 5x + 3e3x.

Each k multiplies out front and stays put inside; the sign change on cosine's derivative is the one detail that separates these five marks.

The logarithm curve with short tangent segments at x equal to 1, 2 and 4, of slopes 1, one half and one quarter: the slope of ln x at every point is exactly 1 over xslope 1 at x = 1slope ½ at x = 2slope ¼ at x = 4y = ln xthe slope readout of ln x is 1/x
FIG. 2ln x with tangents at 1, 2 and 4: slopes 1, ½ and ¼. The logarithm's derivative is 1/x, and the curve flattens on schedule.

Why sine's derivative is cosine

The specification asks for sin x from first principles, and the proof is the radians lesson cashing its cheque. The chord gradient expands by the compound formula, and the two small-angle facts finish it.

WORKED EXAMPLE

sin x from first principles

Prove from first principles that the derivative of sin x is cos x, for x in radians.

The chord gradient is [sin (x + h) − sin x]/h = [sin x cos h + cos x sin h − sin x]/h.

Regroup: sin x (cos h − 1)/h + cos x (sin h/h).

As h → 0, the small-angle approximations give (cos h − 1)/h → 0 and sin h/h → 1, leaving cos x. ∎

Every ingredient was named: compound formula, then the two limits. In degrees, sin h/h approaches π/180 instead of 1, the derivative gains that factor, and the opening lie falls with it.

YOUR TURN

A tangent on a trig curve

Find the equation of the tangent to y = sin 2x at the point where x = π/6, before opening the working.

Show the working

The point: y = sin (π/3) = √3/2.

The gradient: dy/dx = 2 cos 2x, which at x = π/6 is 2 cos (π/3) = 1.

Tangent: y − √3/2 = 1 × (x − π/6), that is y = x − π/6 + √3/2.

Exact values carried the whole question; the only calculus was one shelf lookup and one k multiplying out.

TRY IT UNSEEN

A base that is not e

Find the gradient of y = 5 × 2x at x = 3.

Show the working

The shelf gives dy/dx = 5 × 2x ln 2.

At x = 3 the gradient is 5 × 8 × ln 2 = 40 ln 2 ≈ 27.7.

The ln 2 is what distinguishes base 2 from base e; only e's curve grows at exactly its own height, as the exponentials lesson promised, and every other base carries its logarithm as a correction factor.

THE EXAM BIT

  • Radians throughout; the trig derivatives are false in degrees and the mark scheme knows it.
  • The k multiplies out front and survives inside: sin 3x gives 3 cos 3x, both threes present.
  • Cosine's derivative carries the minus sign; write the shelf line before substituting anything.
  • First-principles proofs for sin x want the compound expansion, the regrouping, and both small-angle limits named.
  • For akx, the derivative is kakx ln a; forgetting the ln a is the standard lost mark.

CHECK YOURSELF

Differentiate y = 3 ln x − cos 4x, and find the gradient at x = π/4.

Show a hint

Termwise from the shelf; cos π has an exact value.

Show the answer

dy/dx = 3/x + 4 sin 4x.

At x = π/4: 3/(π/4) + 4 sin π = 12/π + 0 = 12/π ≈ 3.82.

The sine term vanished at a multiple of π, which exact-value fluency spots before the calculator comes out.

Sine to cosine, cosine to minus sine, e to itself, ln to the reciprocal: the shelf, in radians.

Every k multiplies out front; every base other than e pays a factor of its log.

CHECK YOUR PROGRESS

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  • Differentiate sin kx, cos kx, tan kx, ekx, akx and ln x fluently.
  • Differentiate sin x from first principles using the small-angle approximations.
  • Find gradients and tangents on trig, exponential and log curves.

No animated video for this topic yet; these notes stand alone.