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

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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. 

Let start with sin x.

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:

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

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

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

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

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

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


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