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Differentiation

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Question 43

Use the derivatives of sin⁡(x)\sin(x)sin(x) and cos⁡(x)\cos(x)cos(x) to show that:

a.

ddx(tan⁡x)=ddx(sin⁡xcos⁡x)=sec⁡2x\displaystyle \frac{d}{dx}(\tan x)=\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)=\sec^2 xdxd​(tanx)=dxd​(cosxsinx​)=sec2x

[2]
b.

ddx(sec⁡x)=ddx(1cos⁡x)=sec⁡xtan⁡x\displaystyle \frac{d}{dx}(\sec x)=\frac{d}{dx}\left(\frac{1}{\cos x}\right)=\sec x\tan xdxd​(secx)=dxd​(cosx1​)=secxtanx

[2]
c.

ddx(cot⁡x)=ddx(cos⁡xsin⁡x)=−cosec2x\displaystyle \frac{d}{dx}(\cot x)=\frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)=-\text{cosec}^2 xdxd​(cotx)=dxd​(sinxcosx​)=−cosec2x

[2]
d.

ddx(cosec x)=ddx(1sin⁡x)=−cosec xcot⁡x\displaystyle \frac{d}{dx}(\text{cosec }x)=\frac{d}{dx}\left(\frac{1}{\sin x}\right)=-\text{cosec }x\cot xdxd​(cosec x)=dxd​(sinx1​)=−cosec xcotx

[2]
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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