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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.

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

Dr Brian Brooks
Mathematics InSight

Watch this video to figure out the differential of \(\cos x\).

First of all, what is the relationship of the yellow curve to the white graph \(y=\cos x\)?

The yellow curve represents the gradient of the white curve.

Use the equation of the yellow curve to find 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 \[\frac{\mathrm{d}}{\mathrm{d}x}\cos x=-\sin x\]

Use the definition of a differential to find the differential of \(\cos x\):

\(f(x)\) \(=\) \(\cos x\)
\(f'(x)\) \(=\) \(\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\cos(x+h)-\cos x}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h}\)
\(=\) \(\displaystyle\lim_{h\to 0}\dfrac{-\cos x(1-\cos h)-\sin x\sin h}{h}\)
\(=\) \(-\cos x\displaystyle\lim_{h\to 0}\dfrac{1-\cos h}{h}-\sin x\lim_{h\to 0}\dfrac{\sin h}{h}\)
\(=\) \(-\sin x\)
This video demonstrates the differential of \(\tan x\).
Use the quotient rule to find the differential of \(\tan x\).
\(f(x)\) \(=\) \(\tan x=\dfrac{\sin x}{\cos x}\)
\(f'(x)\) \(=\) \(\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}\)
\(=\) \(\dfrac{1}{\cos^2 x}=\sec^2 x\)
Have a go at a first-principles differential of \(\tan x\).
\(\dfrac{\tan(x+h)-\tan x}{h}\) \(=\) \(\dfrac{1}{h}\!\left[\dfrac{\tan x+\tan h}{1-\tan x\tan h}-\tan x\right]\)
\(=\) \(\dfrac{1}{h}\dfrac{\tan x+\tan h-\tan x+\tan^2 x\tan h}{1-\tan x\tan h}\)
\(=\) \(\dfrac{1}{h}\dfrac{\tan h(1+\tan^2 x)}{1-\tan x\tan h}\)
\(=\) \(\dfrac{\tan h}{h}\dfrac{1+\tan^2 x}{1-\tan x\tan h}\)
\(=\) \(\dfrac{\tan h}{h}\dfrac{\sec^2 x}{1-\tan x\tan h}\)
\(\to\) \(1\times\sec^2 x\)
\(\Rightarrow\) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\tan x=\sec^2 x\)
And here is the differential of \(\sec x\).
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

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