Differentiation

Further Maths · Further Pure Mathematics 2. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

This topic adds new functions and new methods to A Level differentiation. The hyperbolic functions differentiate almost like the trigonometric ones, but cosh x gives +sinh x with no minus sign. The inverse trigonometric and inverse hyperbolic functions give algebraic results with square roots.

You must also find the second derivative d²y/dx² when the curve is given implicitly or by a parameter. For a parameter t, differentiate dy/dx with respect to t and then divide by dx/dt.

A Maclaurin series writes a function as a polynomial in x, using the values of the function and its derivatives at x = 0.

Rules to remember

d/dx: sinh x → cosh x; cosh x → sinh x; tanh x → sech² x; sech x → −sech x tanh x
d/dx: sin⁻¹ x → 1/√(1 − x²); cos⁻¹ x → −1/√(1 − x²)
d/dx: sinh⁻¹ x → 1/√(x² + 1); cosh⁻¹ x → 1/√(x² − 1); tanh⁻¹ x → 1/(1 − x²)
With ax in place of x use the chain rule, e.g. d/dx sinh⁻¹(ax) = a/√(a²x² + 1)
Parametric: dy/dx = (dy/dt) ÷ (dx/dt), and d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt)
Implicit: differentiate the whole equation twice with respect to x, using d/dx(y²) = 2y(dy/dx), then substitute the point and the value of dy/dx
Maclaurin: f(x) = f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x³/3! + …

Common mistake

For a parametric curve students write d²y/dx² as (d²y/dt²) ÷ (d²x/dt²). This is wrong. Differentiate dy/dx with respect to t, then divide by dx/dt.

Worked example

A curve has parametric equations x = t³, y = t² for t > 0. Find d²y/dx² when t = 1.

dx/dt = 3t² and dy/dt = 2t, so dy/dx = 2t/(3t²) = 2/(3t).
d/dt (dy/dx) = −2/(3t²).
d²y/dx² = [−2/(3t²)] ÷ (3t²) = −2/(9t⁴).
At t = 1: −2/9.

Answer: d²y/dx² = −2/(9t⁴) = −2/9 when t = 1

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Power Series (Maclaurin)

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