←

More differentials of circular functions

We've explored sine, cosine, and tangent functions in some depth. Now it's time to broaden our horizons with some new functions.

How to Navigate:

  • Use the arrow keys on your keyboard to move between pages
  • Use swiping gestures on touch devices
  • Hover near the bottom of the page to reveal navigation buttons

Take your time with each video. Pause, rewind, and think before moving on!

Dr Brian Brooks
Mathematics InSight

What is the differential of \(\cos x\) ?

It's easy to see from the graph \(y=\cos x\) and the fact that it is a translation of the sin graph, that \[\frac{\mathrm{d}}{\mathrm{d}x}\cos x=-\sin x\]

but we can also use the definition of a differential:

\(f(x)\) \(=\) \(\cos x\)
\(f'(x)\) \(=\) \(\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\cos(x+h)-\cos x}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{-\cos x(1-\cos h)-\sin x\sin h}{h}\)
\(=\) \(-\cos x\displaystyle\lim_{h\to 0}\dfrac{1-\cos h}{h}-\sin x\lim_{h\to 0}\dfrac{\sin h}{h}\)
\(=\) \(-\sin x\)
Use the quotient rule to find the differential of \(\tan x\).
\(f(x)\) \(=\) \(\tan x=\dfrac{\sin x}{\cos x}\)
\(f'(x)\) \(=\) \(\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}\)
\(=\) \(\dfrac{1}{\cos^2 x}=\sec^2 x\)
\(\dfrac{\tan(x+h)-\tan x}{h}\) \(=\) \(\dfrac{1}{h}\!\left[\dfrac{\tan x+\tan h}{1-\tan x\tan h}-\tan x\right]\)
\(=\) \(\dfrac{1}{h}\dfrac{\tan x+\tan h-\tan x+\tan^2 x\tan h}{1-\tan x\tan h}\)
\(=\) \(\dfrac{1}{h}\dfrac{\tan h(1+\tan^2 x)}{1-\tan x\tan h}\)
\(=\) \(\dfrac{\tan h}{h}\dfrac{1+\tan^2 x}{1-\tan x\tan h}\)
\(=\) \(\dfrac{\tan h}{h}\dfrac{\sec^2 x}{1-\tan x\tan h}\)
\(\to\) \(1\times\sec^2 x\)
\(\Rightarrow\) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\tan x=\sec^2 x\)
Use the quotient rule to find the differential of \(\sec x\).
\(f(x)\) \(=\) \(\sec x=\dfrac{1}{\cos x}\)
\(f'(x)\) \(=\) \(\dfrac{\sin x}{\cos^2 x}=\sec x\tan x\)

or if you prefer to differentiate everything using only the essential definition of a differential:

\[\begin{aligned}f(x)&=\sec x\end{aligned}\]
\(f'(x)\) \(=\) \(\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\sec(x+h)-\sec x}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\dfrac{1}{\cos(x+h)}-\dfrac{1}{\cos x}}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\cos x-\cos(x+h)}{h\cos x\cos(x+h)}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\cos x-\cos(x+h)}{h}\cdot\dfrac{1}{\cos x\cos(x+h)}\)
\(=\) \(\sin x\cdot\dfrac{1}{\cos^2 x}\)
\(=\) \(\dfrac{\sin x}{\cos^2 x}=\sec x\tan x\)
Use the quotient rule to find the differential of \(\operatorname{cosec} x\).
\(f(x)\) \(=\) \(\operatorname{cosec} x=\dfrac{1}{\sin x}\)
\(f'(x)\) \(=\) \(-\dfrac{\cos x}{\sin^2 x}=-\operatorname{cosec} x\cot x\)
Use the quotient rule to find the differential of \(\cot x\).
\(f(x)\) \(=\) \(\cot x=\dfrac{\cos x}{\sin x}\)
\(f'(x)\) \(=\) \(\dfrac{-\sin^2 x-\cos^2 x}{\sin^2 x}\)
\(=\) \(-\dfrac{1}{\sin^2 x}=-\operatorname{cosec}^2 x\)

Well done!

You've found the differentials of the five reciprocal circular functions: \(\cos x\), \(\tan x\), \(\sec x\), \(\operatorname{cosec} x\), and \(\cot x\).

What's Next:

  • The next worksheet covers the differentials of the inverse circular functions
  • You'll use implicit differentiation to find the derivatives of arcsin, arccos, arctan, and more

Dr Brian Brooks
Mathematics InSight

Page of 7