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}