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Differentiation Techniques Notes — JC H2

Differentiation Techniques covers how to differentiate a wide range of functions: powers and roots, exponentials and logarithms, trigonometric and inverse trigonometric functions, and combinations of these. It also covers curves defined implicitly, where dy/dx is found in terms of x and y, and curves defined by parametric equations.

This set is the Chapter 10 assignment of the 2026 PU2 H2 Mathematics course at Millennia Institute. Its eight questions, with answers, move from direct differentiation and second derivatives to sketching parametric curves, finding where the gradient is zero and finding unknown constants from given gradients.

What these notes cover

  • •Powers, roots and logarithms
  • •Exponential and trigonometric functions
  • •Inverse trigonometric functions
  • •Second derivatives
  • •Implicit differentiation
  • •Exponents involving x, e.g. 5^(√x) and (x² + x)^x
  • •Parametric equations to Cartesian form
  • •Gradient of a parametric curve
  • •Sketching a parametric curve
  • •Points where the gradient is zero
  • •Intersection of a line and a parametric curve
  • •Finding unknown constants from gradients

Key formulas

Logarithm of sec x + tan x

ddxln⁡(sec⁡x+tan⁡x)=sec⁡x\frac{d}{dx}\ln(\sec x + \tan x) = \sec x

Chain rule on a logarithm; the result simplifies to a single term.

Logarithm of sin²x

ddxln⁡(sin⁡2x)=2cot⁡x\frac{d}{dx}\ln(\sin^2 x) = 2\cot x

Chain rule; the answer is written in terms of cot x.

Inverse sine with a scaled argument

ddxsin⁡−1(x10)=1100−x2\frac{d}{dx}\sin^{-1}\left(\frac{x}{10}\right) = \frac{1}{\sqrt{100 - x^2}}

Inverse trigonometric function of x ÷ 10.

Product of exponential and trigonometric functions

ddx[e3xsin⁡22x]=3e3xsin⁡22x+4e3xsin⁡2xcos⁡2x\frac{d}{dx}\left[e^{3x}\sin^2 2x\right] = 3e^{3x}\sin^2 2x + 4e^{3x}\sin 2x\cos 2x

Product rule, with the chain rule for e^(3x) and for sin²2x.

Exponential with base 5

ddx(5x)=5xln⁡52x\frac{d}{dx}\left(5^{\sqrt{x}}\right) = \frac{5^{\sqrt{x}}\ln 5}{2\sqrt{x}}

Base 5 with exponent √x; the factor ln 5 comes from the base.

Variable base and power

ddx[(x2+x)x]=(x2+x)x[2x+1x+1+ln⁡(x2+x)]\frac{d}{dx}\left[(x^2+x)^x\right] = (x^2+x)^x\left[\frac{2x+1}{x+1} + \ln(x^2+x)\right]

Both the base and the exponent contain x.

Worked example

For y = ln(x ÷ (x + 1)), find and simplify dy/dx and hence find d²y/dx².

  1. 1Split the logarithm first: y = ln x − ln(1 + x).
  2. 2Differentiate each term: dy/dx = 1 ÷ x − 1 ÷ (x + 1).
  3. 3Differentiate again: d²y/dx² = −1 ÷ x² + 1 ÷ (x + 1)².

Questions students ask

What is the derivative of ln(sec x + tan x)?

d/dx [ln(sec x + tan x)] = sec x. In the same set, ln(sin²x) differentiates to 2 cot x and sin⁻¹(x ÷ 10) differentiates to 1 ÷ √(100 − x²).

How do you differentiate 5 to the power of root x?

d/dx [5^(√x)] = (5^(√x) ln 5) ÷ (2√x). A related question differentiates (x² + x)^x, where both the base and the power contain x, giving (x² + x)^x [(2x + 1) ÷ (x + 1) + ln(x² + x)].

How do you find dy/dx in terms of x and y?

This is needed when a curve is given implicitly, such as x²y + xy³ = 32. Differentiating gives dy/dx = −(2xy + y³) ÷ (x² + 3xy²), which contains both x and y.

How do you find the Cartesian equation of a parametric curve?

Remove the parameter to get a relation between x and y only. For x = ln t and y = t + 10 with t ≥ 1, the Cartesian equation is y = eˣ + 10, x ≥ 0; for x = 3 ÷ t² and y = 3t with t ≠ 0, it is xy² = 27. Both curves are also sketched using a graphic calculator.

In the full PDF

  • •All eight questions with answers, including every part of Question 1
  • •Quotient-rule and inverse cosine questions, such as e^(x²) ÷ ln x
  • •Implicit differentiation with exponential and tan⁻¹ terms
  • •Gradient of a parametric curve at (1, 1) without a calculator, plus a 'show that' in θ
  • •A parametric curve to sketch with intercepts and endpoints, its zero-gradient point and its intersection with a line
  • •Finding a and b in y = ax² + b ÷ x from the gradients at two points