Use the derivatives of sin(x)\sin(x)sin(x) and cos(x)\cos(x)cos(x) to show that:
ddx(tanx)=ddx(sinxcosx)=sec2x\displaystyle \frac{d}{dx}(\tan x)=\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)=\sec^2 xdxd(tanx)=dxd(cosxsinx)=sec2x
ddx(secx)=ddx(1cosx)=secxtanx\displaystyle \frac{d}{dx}(\sec x)=\frac{d}{dx}\left(\frac{1}{\cos x}\right)=\sec x\tan xdxd(secx)=dxd(cosx1)=secxtanx
ddx(cotx)=ddx(cosxsinx)=−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
ddx(cosec x)=ddx(1sinx)=−cosec xcotx\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
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.