Differentiating \(sin(x)\) and \(cos(x)\)
Differentiating sin(x) and cos(x)
- Using small angle approximations:
- \(\sin(x)=x\)
- \(\cos(x)=1-\frac{1}{2}x^2\)
- This means that \(\lim_{ h \to 0 }\frac{\sin(h)}{h}=\lim_{ h \to 0 }\frac{h}{h}=1\)
- and \(\lim_{ h \to 0 }\frac{\cos h-1}{h}=\lim_{ h \to 0 }\frac{1-\frac{1}{2}h^2-1}{h}=\lim_{ h \to 0 }-\frac{1}{2}h=0\)
\(\sin(x)\)
- Let \(f(x)=\sin(x)\)
- \(f^{'}(x)=\lim_{ h \to 0 }\frac{f(x+h)-f(x)}{h}\)
- \(=\lim_{ h \to 0 }\frac{\sin(x+h)-\sin(x)}{h}\)
- \(=\lim_{ h \to 0 }\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}\)
- \(=\lim_{ h \to 0 }\frac{\sin x(\cos h-1)}{h}+\frac{\cos x\sin h}{h}\)
- \(=\lim_{ h \to 0 }\frac{\sin x\left( 1-\frac{h^2}{2}-1 \right)}{h}+\frac{\cos x\times h}{h}\)
- \(=\lim_{ h \to 0 }\frac{h}{2}\sin x+\cos x\)
- \(=\cos x\)
\(\cos (x)\)
- Let \(f(x)=\sin(x)\)
- \(f^{'}(x)=\lim_{ h \to 0 }\frac{f(x+h)-f(x)}{h}\)
- \(f^{'}(x)=\lim_{ h \to 0 }\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\)
- \(=\lim_{ h \to 0 }\frac{\cos x(\cos h-1)}{h}-\frac{\sin x\sin h}{h}\)
- \(=\lim_{ h \to 0 }\frac{\cos(x)\left( 1-\frac{h^2}{2}-1 \right)}{h}-\frac{\sin x \times h}{h}\)
- \(=\lim_{ h \to 0 }-\frac{h}{2}\cos x-\sin x\)
- \(=-\sin x\)
If there is a \(k\) in front...
- \(\frac{d}{dx}(\sin kx)=k\cos kx\)
- \(\frac{d}{dx}(\cos kx)=-k\sin kx\)
Examples
- \(\frac{d}{dx}(\sin3x)=3\cos3x\)
- \(\frac{d}{dx}(5x)=-5\sin 5x\)
- \(\frac{d}{dx}(3\sin 5x)=3(5\cos5x)=15\cos 5x\)
- \(\frac{d}{dx}\left( -\frac{2}{3}\cos\frac{1}{2}x \right)=-\frac{2}{3}\left( -\frac{1}{2}\cos\frac{1}{2}x \right)=\frac{1}{3}\cos\frac{1}{2}x\)
Finding stationary points
Find the stationary points of \(y=\frac{1}{2}x-\cos 2x, 0\leq x\leq \pi\)
\(\frac{d}{dx}=\frac{1}{2}+2\sin2x\)
\(\frac{1}{2}+2\sin2x=0\)
\(2\sin2x=-\frac{1}{2}\)
\(\sin2x=-\frac{1}{4}\)
\(2x=\sin^{-1}\left( -\frac{1}{4} \right)\)
\(2x=-14.477\)
\(x=7.238\)
Exercise 9a
1a.
- Let \(f(x)=\sin(x)\)
- \(f^{'}(x)=\lim_{ h \to 0 }\frac{f(x+h)-f(x)}{h}\)
- \(f^{'}(x)=\lim_{ h \to 0 }\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\)
- \(=\lim_{ h \to 0 }\frac{\cos x(\cos h-1)}{h}-\frac{\sin x\sin h}{h}\)
- \(=\lim_{ h \to 0 }\frac{\cos h-1}{h}\cos x-\frac{\sin h}{h}\sin x\)
1b.
- \(=\lim_{ h \to 0 }\frac{\cos h-1}{h}\cos x-\frac{\sin h}{h}\sin x\)
- \(=\lim_{ h \to 0 }\frac{1-\frac{h^2}{2}-1}{h}\cos x-\frac{h}{h}\sin x\)
- \(=\lim_{ h \to 0 }-\frac{h}{2}\cos x-\sin x\)
- \(=-\sin x\)
- a) \(\frac{dy}{dx}2\cos x=-2\sin x\)
b) \(\frac{dy}{dx}2\sin \frac{1}{2}x=\cos\frac{1}{2}x\)
c) \(\frac{dy}{dx}\sin8x=8\cos8x\)
d) \(\frac{dy}{dx}6\sin\frac{2}{3}x=4\cos\frac{2}{3}x\)
- a) \(\frac{dy}{dx}2\cos x=-2\sin x\)
b) \(\frac{dy}{dx}6\cos\frac{5}{6}x=-5\sin\frac{5}{6}x\)
c) \(\frac{dy}{dx}4\cos\frac{1}{2}x=-2\sin\frac{1}{2}x\)
d) \(\frac{dy}{dx}3\cos2x=-6\sin2x\)
- a) \(\frac{dy}{dx}\sin2x+\cos3x=2\cos 2x-\sin 3x\)
b) \(\frac{dy}{dx}2\cos4x-4\cos x+2\cos7x=-8\sin4x+4\sin x-14\sin7x\)
c) \(\frac{dy}{dx}x^2+4\cos3x=2x-12\sin3x\)
d) \(\frac{dy}{dx}\frac{1}{x}+x\sin5x=-\frac{1}{x^2}+10x\cos5x\)
- \(\frac{dy}{dx}x-\sin3x=1-3\cos3x\)
\(1-3\cos3x=0\)
\(\cos3x=\frac{1}{3}\)
\(x=0.41\)
\(x=1.68\)
\(x=2.50\)
- \(\frac{dy}{dx}2\sin4x-4\cos2x\)
\(=8\cos4x+8\sin2x\)
\(=8\cos2\pi+8\sin \pi\)
\(=8+0\)
\(=8\)
Equations of Tangents
Find the equation of the tangent to \(y=2\sin(2x)\) at the point where \(x=\frac{\pi}{6}\)
- \(2\sin \frac{\pi}{3}=\sqrt{ 3 }\)
- Co-ordinates are \(\left( \frac{\pi}{6},\sqrt{ 3 } \right)\)
- \(\frac{dy}{dx}=4\cos(2x)\)
- gradient: \(4\cos \frac{\pi}{3}=2\)
- \(y-\sqrt{ 3 }=2\left( x-\frac{\pi}{6} \right)\)
- \(y-\sqrt{ 3 }=2x-\frac{\pi}{3}\)
- \(y=2x+\sqrt{ 3 }+\frac{\pi}{3}\)