Before we start, recall that \[\lim_{x\to0}\frac{\sin x}{x}=0\tag{a}\] also
\[\begin{align} \lim_{x\to0}\frac{\cos x-1}{x} & =\lim_{x\to0}\frac{2\sin^{2}\frac{x}{2}}{x}\\ & =\lim_{x\to0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\sin\frac{x}{2}\right)\\ & =\lim_{x\to0}\frac{\sin\frac{x}{2}}{\frac{x}{2}}\times\lim_{x\to0}\sin\frac{x}{2}\\ & =\lim_{u\to0}\frac{\sin u}{u}\times\lim_{x\to0}\sin u\tag{$\textrm{let } u=\frac{x}{2}$}\\ & =1\times0\\ & =0.\tag{b}\end{align}\]
and
\[\begin{align} \sin(A+B) & =\sin A\cos B+\cos A\sin B\\ \cos(A+B) & =\cos A\cos B-\sin A\sin B.\end{align}\]

The derivative of \(y=\sin x\)

\[\begin{align} \frac{dy}{dx} & =\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}\\ & =\lim_{h\to0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ & =\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}\right)\lim_{h\to0}\left(\cos x\frac{\sin h}{h}\right)\\ & =\sin x\times\lim_{h\to0}\frac{\cos h-1}{h}+\cos x\times\lim_{h\to0}\frac{\sin h}{h}\\ & =\sin x\times0+\cos x\times1\\ & =\cos x.\end{align}\]

Therefore

\[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\sin x=\cos x.\tag{c}}\]

The derivative \(y=\cos x\)

\[\begin{align} \frac{dy}{dx} & =\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}\\ & =\lim_{h\to0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\ & =\lim_{h\to0}\frac{\cos x(\cos h-1)-\sin x\sin h}{h}\\ & =\left(\lim_{h\to0}\frac{\cos x(\cos h-1)}{h}\right)-\left(\lim_{h\to0}\frac{\sin x\sin h}{h}\right)\\ & =\cos x\left(\lim_{h\to0}\frac{\cos h-1}{h}\right)-\sin x\left(\lim_{h\to0}\frac{\sin h}{h}\right)\\ & =\cos x\times0-\sin x\times1\\ & =-\sin x\end{align}\]

Therefore
\[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\cos x=-\sin x.\tag{d}}\]

  • \(\sin x\) and \(\cos x\) are differentiable everywhere
  • You need to memorize formulas (c) and (d)

The derivative of \(y=\tan x\)

To differentiate \(y=\tan x\), we write \(\tan x\) as \(\frac{\sin x}{\cos x}.\) Then we use \[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u’v-v’x}{v^{2}}.\] Thus

\[\begin{align} \frac{d}{dx}\tan x & =\frac{d}{dx}\frac{\sin x}{\cos x}\\ & =\frac{\cos x\cos x-(-\sin x)\sin x}{\cos^{2}x}\\ & =\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}\\ & =\frac{1}{\cos^{2}x}\\ & =\sec^{2}x.\end{align}\]
That is, \[\frac{d}{dx}\tan x=\sec^{2}x.\] Also we recall that
\[\frac{1}{\cos^{2}x}=\frac{\sin^{2}+\cos^{2}x}{\cos^{2}x}=\tan^{2}x+1.\]
Therefore, we can also write
\[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\tan x=\sec^{2}x=1+\tan^{2}x.}\]

The derivative of \(y=\cot x\)

Similar to \(y=\tan x,\) we can differentiate \(y=\cot x.\)
\[\begin{align} \frac{d}{dx}\cot x & =\frac{d}{dx}\frac{\cos x}{\sin x}\\ & =\frac{-\sin x\times\sin x-\cos x\times\cos x}{\sin^{2}x}\\ & =-\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x}\\ & =-\frac{1}{\sin^{2}x}\\ & =-\csc^{2}x.\end{align}\]

\[\frac{d}{dx}\cot x=-\csc^{2}x\] Because \[\frac{1}{\sin^{2}x}=\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x}=1+\cot^{2}x,\] we have
\[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\cot x=-\csc^{2}x=-(1+\cot^{2}x).}\]

The derivative of \(y=\sec x\)

\[\begin{align} \frac{d}{dx}\sec x & =\frac{d}{dx}\frac{1}{\cos x}\\ & =\frac{0\times\cos x-(-\sin x)\times1}{\cos^{2}x}\\ & =\frac{\sin x}{\cos^{2}}\\ & =\frac{\sin x}{\cos x}\frac{1}{\cos x}\\ & =\tan x\ \sec x\end{align}\]

The derivative of \(y=\csc x\)
\[\begin{align} \frac{d}{dx}\csc x & =\frac{d}{dx}\frac{1}{\sin x}\\ & =\frac{0\times\sin x-\cos x\times1}{\sin^{2}x}\\ & =\frac{-\cos x}{\sin^{2}x}\\ & =-\frac{\cos x}{\sin x}\frac{1}{\sin x}\\ & =-\cot x\ \csc x.\end{align}\]

Tip for memorizing these formulas

  • The three “co-” functions (cosine x, cotangent x, and cosecant x) have minus sign in their derivatives.
Example 1

Differentiate (a) \(y=\sin ax\) (b) \(y=\cos ax\) (c) \(y=\tan ax\)

Solution

(a) Let \(u=ax\) . Thus \(y=\sin u\) and
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =\cos u\times a\\ & =a\cos ax\end{align}\]
(b) Similar to (a), let \(u=ax\) and \(y=\cos u\). Therefore,
\[\begin{align} \frac{dy}{dx} & =-\sin u\times a\\ & =-a\sin ax.\end{align}\]
(c) Similar to (a) and (b), let \(u=ax\) and \(y=\tan u\). We can show
\[\begin{align} \frac{dy}{dx} & =\left(\frac{d}{du}\tan u\right)\left(\frac{d}{dx}(ax)\right)\\ & =a(1+\tan^{2}ax).\end{align}\]

Example 2

Differentiate \(y=\sin^{2}x\).

Solution

Note that \(y=\sin^{2}x=(\sin x)^{2}.\)

Method 1: We write \(y=u^{2}\) with \(\underline{u=\sin{x}}\)

\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =2u\cos x\\ & =2\sin x\cos x.\end{align}\]
Method 2: We write \(y=\sin^{2}x=\frac{1}{2}(1-\cos2x)\).
\[\begin{align} \frac{dy}{dx} & =-\frac{1}{2}(-2\sin2x)\\ & =\sin2x.\end{align}\]
Note that the results of both methods are the same, as \[\sin2x=2\sin x\cos x.\]

Example 3

Differentiate \(y=\sin^{2}\sqrt{x}\).

Solution

Let \(y=u^{2}\), \(u=\sin v\), and \(v=\sqrt{x}\). Using the chain rule, we have
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}\\ & =(2u)(\cos v)\left(\frac{1}{2\sqrt{x}}\right)\\ & =(2\sin v)(\cos\sqrt{x})\left(\frac{1}{2\sqrt{x}}\right)\\ & =(2\sin\sqrt{x})(\cos\sqrt{x})\left(\frac{1}{2\sqrt{x}}\right).\end{align}\]
We can make it concise by using the identity
\(\sin2\theta=2\sin\theta\cos\theta\):
\[\frac{dy}{dx}=\frac{\sin(2\sqrt{x})}{2\sqrt{x}}.\]

Example 4

If \(z=\cos(\sin^{3}x^{2})\), find \(dz/dx\).

Solution

Let \(z=\cos u,\ u=t^{3},\ t=\sin w,\) and \(w=x^{2}.\) Then
\[\begin{align} \frac{dz}{dx} & =\frac{dz}{du}\frac{du}{dt}\frac{dt}{dw}\frac{dw}{dx}\\ & =-\sin u\times3t^{2}\times\cos w\times2x\end{align}\]
Now we write \(u,t,\) and \(w\) in term of \(x\):
\[\begin{align} \frac{dz}{dx} & =-\sin t^{3}\times3\sin^{2}w\times\cos x^{2}\times2x\\ & =-6\sin(\sin^{3}w)\times\sin^{2}x^{2}\times\cos x^{2}\times x\\ & =-6x(\cos x^{2})(\sin^{2}x^{2})\sin(\sin^{3}x^{2}).\end{align}\]

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