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Differentiation

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Question 81

By using the identities tan⁡θ=sin⁡θcos⁡θ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}tanθ=cosθsinθ​ and sec⁡θ=1cos⁡θ\displaystyle \sec \theta = \frac{1}{\cos \theta}secθ=cosθ1​, and the standard derivatives of sin⁡θ \sin \theta\,sinθ and cos⁡θ\cos \thetacosθ, prove the following results:

a.
ddθ(tan⁡θ)=sec⁡2θ \frac{d}{d\theta}(\tan \theta) = \sec^2 \theta dθd​(tanθ)=sec2θ
[3]
b.
ddθ(sec⁡θ)=sec⁡θtan⁡θ \frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta dθd​(secθ)=secθtanθ
[3]
c.
ddθ(cot⁡θ)=−csc⁡2θ \frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta dθd​(cotθ)=−csc2θ
[3]
d.
ddθ(csc⁡θ)=−csc⁡θcot⁡θ \frac{d}{d\theta}(\csc \theta) = -\csc \theta \cot \theta dθd​(cscθ)=−cscθcotθ
[3]
Markscheme

Differentiation Questions

  1. A Level
  2. /Maths
  3. /Differentiation

717 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.

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