# PROOF OF THE DERIVATIVES OF SIN, COS, TAN, CSC, SEC AND COT

The standard derivative of Sine, cosine, tangent, cosecant, secant, and cotangent is used to differentiate trigonometric functions.

These are outlined below.

$$\frac{d}{dx}(\sin x)=\cos x$$

$$\frac{d}{dx}(\cos x)=-\sin x$$

$$\frac{d}{dx}(\tan x)=-\sec^2 x$$

$$\frac{d}{dx}(\sec x)=\sec x \tan x$$

$$\frac{d}{dx}(\csc x)=-\csc x \cot x$$

$$\frac{d}{dx}(\cot x)=-\csc^2 x$$

But, how do we obtain this derivative? Is it a random guess? I will prove the six derivatives of trigonometric function in this post.

## Prove That The Derivative Of Sin X Is Cos X

Let $f(x)=\sin x$

Recall from the definition of the derivative of a function that

$$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

If $f(x)=\sin x$, then, $f(x+h)=\sin(x+h)$ and

$$f'(x)=\lim_{h \to 0}\frac{\sin(x+h)-\sin(x)}{h}$$

Remember the trig identity that says, $\sin ( a+b)=\sin a \cos b+\cos a \sin b$, therefore, $\sin(x+h)=\sin x \cos h+\cos x \sin h$ and

$$f'(x)=\lim_{h \to 0}\frac{\sin x \cos h+\cos x \sin h-\sin x}{h}$$

Let's do a bit of re-arranging

$$f'(x)=\lim_{h \to 0}\frac{\sin x \cos h-\sin x+\cos x \sin h}{h}$$

$$f'(x)=\lim_{h \to 0}\frac{\sin x(\cos h-1)+\cos x \sin h}{h}$$

Splitting the fraction

$$f'(x)=\lim_{h \to 0}\frac{\sin x(\cos h-1)}{h}+\frac{\cos x \sin h}{h}$$

Now, we turn to the limit expression as we used the sum rule of limits.

$$f'(x)=\lim_{h \to 0}\frac{\sin x (\ cos h-1)}{h}+\lim_{h \to 0}\frac{\cos x \sin h}{h}$$

We can bring sin x and cos x outside the limits because they are functions of x, not h.

$$f'(x)=\sin x\left(\lim_{h \to 0}\frac{\cos h-1}{h}\right)+\cos x\left(\lim_{h \to 0}\frac{\sin h}{h}\right)$$

Now, let use the l'hopital rule which says we should differentiate the numerator and the denominator

$$f'(x)=\sin x\left(\lim_{h \to 0}\frac{-\sin h}{1}\right)+\cos x\left(\lim_{h \to 0}\frac{\cos h}{1}\right)$$

By direct substitution

$$f'(x)=(\sin x)(-\sin 0)+(\cos x)(\cos 0)$$

$$f'(x)=(\sin x)(0)+(\cos x)(1)$$

$$f'(x)=0+\cos x$$

$$f'(x)=\cos x$$

Therefore, the derivative of sine is cosine.

## Prove That The Derivative Of Cos X Is -sin X

Let $f(x)=\cos x$

The definition of the derivative of a function says that

$$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

If $f(x)=\cos x$, then, $f(x+h)=\cos(x+h)$ and

$$f'(x)=\lim_{h \to 0}\frac{\cos(x+h)-\cos x}{h}$$

Recall the trig identity that says $\cos (x+h)=\cos x \cos h-\sin x \sin h$. Accordingly,

$f'(x)=\lim_{h \to 0}\frac{\cos x \cos h-\sin x \sin h-\cos x}{h}$

$f'(x)=\lim_{h \to 0}\frac{\cos x \cos h-\cos x-\sin x \sin h}{h}$

$$f'(x)=\lim_{h \to 0}\frac{\cos x(\cos h-1)-\sin x \sin h}{h}$$

$$f'(x)=\lim_{h \to 0}\frac{\cos x(\cos h-1)}{h}-\frac{\sin x \sin h}{h}$$

Now, we use the difference rule of limits

$$f'(x)=\lim_{h \to 0}\frac{\cos x(\cos h-1)}{h}-\lim_{h \to 0}\frac{\sin x \sin h}{h}$$

We can bring cos x and sin x outside the limits because they are functions of x, not h.

$$f'(x)=\cos x \left(\lim_{h \to 0}\frac{\cos h-1}{h}\right)-\sin x \left( \lim_{h \to 0}\frac{\sin h}{h}\right)$$

Applying l'hopital rule

$$f'(x)=\cos x \lim_{h \to 0}\frac{-\sin h}{1}-\sin x \lim_{h \to 0}\frac{\cos h}{1}$$

$$f'(x)=(\cos x)(-\sin 0)-(\sin x) (\cos 0)$$

$$f'(x)=(\cos x)(0)-(\sin x) (1)$$

$$f'(x)=-\sin x$$

Therefore, the derivative of cosine is negative sine

Before we move to tangent, there are a few things to keep in mind.

$$\tan=\frac{\sin x}{\cos x}$$

$$\sec=\frac{1}{\cos x}$$

$$\csc=\frac{1}{\sin x}$$

$$\cot=\frac{\cos x}{\sin x}$$

## Prove That The Derivative Of Tan X Is Sec²x

Recall that  $\tan x=\frac{\sin x}{\cos x}$. Therefore,

$$\frac{d}{dx}\tan x= \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)$$

Since we have a fraction, we use the quotient rule which states that

$$\left(\frac{f}{g}\right)=\frac{f'g-g'f}{(g)^2}$$

Here, f=sin x, g=cos x, f'=cos x, g'=-sin x

$$\frac{d}{dx}\tan x=\frac{(\cos x)(\cos x)-(-\sin x)(\sin x)}{\cos^2 x}$$

$$\frac{d}{dx}\tan x=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}$$

Recall the trig identity that says that $\cos^2 x+\sin^2 x=1$. See proof here

$$\frac{d}{dx}\tan x=\left(\frac{1}{\cos^2 x}\right)$$

$$\frac{d}{dx}\tan x=\left(\frac{1}{\cos x}\right)^2$$

Because $\sec x=\frac{1}{\cos x}$, therefore

$$\frac{d}{dx}\tan x=\sec^2 x$$

Note that that $\sec^2 x$ is the same as $(\sec x)^2$

$$\frac{d}{dx}(\tan x)=\sec^2 x$$

Therefore, the derivative of tangent is the square of secant.

## Prove That The Derivative Of Sec X Is Tanxsecx

Because $\sec x=\frac{1}{\cos x}$. Accordingly,

$$\frac{d}{dx}(\sec x)=\frac{d}{dx}\left(\frac{1}{cos x}\right)$$

Since we have a fraction, we use the quotient rule which is exemplified as:

$$\left(\frac{f}{g}\right)=\frac{f'g-g'f}{(g)^2}$$

Here, f=1, f'=0, g=cos x, g'=-sin x

$$\frac{d}{dx}(\sec x)=\frac{(0)(\cos x)-(-\sin x)(1)}{(\cos x)^2}$$

$$\frac{d}{dx}(\sec x)=\frac{\sin x}{(\cos x)^2}$$

$$\frac{d}{dx}(\sec x)=\frac{\sin x}{\cos x}\times\frac{1}{\cos x}$$

Because $\tan x=\frac{\sin x}{\cos x}$, and $\sec x=\frac{1}{\cos x}$

$$\frac{d}{dx}\sec x=\tan x \sec x$$

Therefore the derivative of secant is simply the multiple of tangent and covenant

## Prove that the Derivative Of Cscx Is -Cscxcotx

As I told you earlier, $\csc x=\frac{1}{\sin x}$. By the same reasoning

$$\frac{d}{dx}\csc x=\frac{d}{dx}\left(\frac{1}{\sin x}\right)$$

Since we have a quotient, we use the quotient rule, which is exemplified as:

$$\left(\frac{f}{g}\right)=\frac{f'g-g'f}{(g)^2}$$

By comparison, f=1, f'=0, g=sin x, g'=cos x

$$\frac{d}{dx}(\csc x)=\frac{0(\sin x)-(\cos x)(1)}{(\sin x)^2}$$

$$\frac{d}{dx}(\csc x)=\frac{-\cos x}{(\sin x)^2}$$

$$\frac{d}{dx}\csc x=\frac{-\cos x}{\sin x}\times\frac{1}{\sin x}$$

Since $\cot x=\frac{\cos x}{\sin x}$ and  $\csc x=\frac{1}{\sin x}$, therefore

$$\frac{d}{dx}\csc x=-\cot x \csc x$$

Therefore, the derivative of cosecant is negative contingent multiplied by cosecant.

Prove That The Derivative Of Cot X Is -csc² X

$\cot x=\frac{\cos x}{\sin x}$. This means

$$\frac{d}{dx}\cot x=\frac{d}{dx}\frac{\cos x}{\sin x}$$

Because cot x is a fraction, we use the quotient rule, which is exemplified as

$$\left(\frac{f}{g}\right)=\frac{f'g-g'f}{(g)^2}$$

Here, f=cos x, f'=-sin x, g=sin x, g'=cos x

$$\frac{d}{dx}(\cot x)=\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{(\sin x)^2}$$

$$\frac{d}{dx}(\cot x)=\frac{-(\sin x)^2-(\cos x)^2}{(\sin x)^2}$$

$$\frac{d}{dx}(\cot x)=\frac{-((\sin x)^2+(\cos x)^2)}{(\sin x)^2}$$

Remember that $(\sin x)^2+(\cos x)^2=1$. See proof here

$$\frac{d}{dx}(\cot x)=\frac{-1}{(\sin x)^2}$$

If $\csc x=\frac{1}{\sin x}$, then

$$\frac{d}{dx}\cot x=-\csc^2$$

Thus, the derivative of tangent is negative the square of cosecant.

Related posts

To repeat, the standard derivative of trigonometric functions are:

$$\frac{d}{dx}(\sin x)=\cos x$$

$$\frac{d}{dx}(\cos x)=-\sin x$$

$$\frac{d}{dx}(\tan x)=-\sec^2 x$$

$$\frac{d}{dx}(\sec x)=\sec x \tan x$$

$$\frac{d}{dx}(\csc x)=-\csc x \cot x$$

$$\frac{d}{dx}(\cot x)=-\csc^2 x$$

Help us grow our readership by sharing this post