Differentiating Trigonometric Functions
Differentiating \(\tan(x)\)
- \(u=\sin x\)
- \(v=\cos x\)
- \(\frac{d}{dx}\frac{u}{v}=\frac{(\cos x\times \cos x)-(\sin x\times-\sin x)}{\cos^2x}\)
- \(=\frac{\cos^2x+\sin^2x}{\cos^2x}\)
- \(=\frac{1}{\cos^2x}=\sec^2x\)
Differentiating \(\sec (x)\)
- \(u=1\)
- \(v=\cos x\)
- \(\frac{d}{dx}\frac{u}{v}=\frac{(0\times \cos x)-(1\times-\sin x)}{\cos^2x}\)
- \(=\frac{\sin x}{\cos^2x}\)
- \(=\frac{\sin x}{\cos x}\times\frac{1}{\cos x}\)
- \(=\sec x\tan x\)
Show that \(\frac{d}{dx}(\sec^{2}3x)\) can be written in the form \(\mu(\tan_{3}x+\tan^{3}3x)\) where \(\mu\) is a constant.
- \(\frac{d}{dx}(\sec^{2}3x)=\frac{dy}{dx}(\sec3x)^2\)
- \(=2\sec3x\times3\sec3x\tan 3x=6\sec^23x\tan3x\)
- \(\sec^2x=1+\tan^2x\)
- So \(6\sec^23x\tan3x=6(1+\tan^{2}3x)(\tan3x)=6\tan3x+6\tan^{3}3x=6(\tan3x+\tan^{3}3x)\)
- So \(\mu=6\)
Show that if \(y=\arcsin x\), then \(\frac{dy}{dx}=\frac{1}{\sqrt{ 1-x^2 }}\)
- \(y=\arcsin x\)
- \(x=\sin y\)
- \(\frac{dx}{dy}=\cos y\)
- \(\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{ 1-\sin^2y }}=\frac{1}{\sqrt{ 1-x^2 }}\)
Given that \(y=\arcsin x^2\) find \(\frac{dy}{dx}\)
- \(y=\arcsin x^2\)
- \(x^2=\sin y\)
- \(x=\sqrt{ \sin y }=(\sin y)^{\frac{1}{2}}\)
- \(\frac{dx}{dy}=\frac{1}{2}(\sin y)^{-\frac{1}{2}}\times \cos y\)
- \(=\frac{\cos y}{2\sqrt{ \sin y }}\)
- \(=\frac{2\sqrt{ \sin y }}{\cos y}\)
- \(=\frac{2x}{\cos y}\)
- \(\frac{dy}{dx}=\frac{2x}{\sqrt{ 1-x^4 }}\)
\(x=\sec2y\) Find \(\frac{dx}{dy}\)
- \(\frac{dx}{dy}=2\sec2y\tan2y\)
- \(\frac{dy}{dx}=\frac{1}{2\sec2y\tan2y}\)
- \(=\frac{1}{\sqrt{ 2x }}\times\frac{1}{\sqrt{ x^2-1 }}\)