# Derivative of cos3x by First Principle

The derivative of cos3x is equal to -3sin3x. Here we learn how to differentiate cos3x and find the derivative of cos3x by first principle.

## Derivative Formula of cos3x

The formula of the derivative of cos3x is given by

d(cos3x)/dx = -3sin3x.

That is, the derivative of cos3x is equal to the negative of three times of sin3x.

## Derivative of cos3x using First Principle

Using the first principle of derivatives, the derivative of a function f(x) is given by the following limit formula:

$\dfrac{d}{dx}\big(f(x) \big)$ = limh→0 $\dfrac{f(x+h)-f(x)}{h}$.

Put f(x)=cos3x. Then the derivative of cos3x using first principle will be equal to

$\dfrac{d}{dx}\big(\cos 3x \big)$ = limh→0 $\dfrac{\cos 3(x+h)- \cos 3x}{h}$

= limh→0 $\dfrac{\cos (3x+3h)- \cos 3x}{h}$

= limh→0 $\dfrac{\cos 3x \cos 3h – \sin 3x \sin 3h – \cos 3x}{h}$. Here we have used the trigonometric formula: cos(A + B) = cosA cosB – sinA sinB.

= limh→0 $\dfrac{\cos 3x(\cos 3h -1) -\sin 3x \sin 3h}{h}$

= limh→0 $\dfrac{\cos 3x(\cos 3h -1)}{h}$ – limh→0 $\dfrac{\sin 3x \sin 3h}{h}$ by the sum rule of limits.

= 3cos3x limh→0 $\dfrac{\cos 3h -1}{3h}$ – sin3x limh→0 $\dfrac{\sin 3h}{h}$

= 3cos3x × 0 – sin3x × 3 by the limit formulas limx→0 $\dfrac{\cos x -1}{x}$ =0 and limx→0 (sinbx/x) = b.

= 0 – 3sin3x

= -3sin3x.

So the derivative of cos3x by first principle is -3sin3x and this is obtained by the first principle of derivatives.

Derivative of tan3x: The derivative of tan3x is 3sec2 3x.

Derivative of sin5x: The derivative of sin5x is 5cos5x.

Derivative of cos2x: The derivative of cos2x is -sin2x.

Derivative of 2x: The derivative of 2x is 2xln2.

## FAQs

### Q1: What is derivative of cos3x?

Answer: The derivative of cos3x is -3sin3x. It is obtained as follows: d(cos3x)/dx = d(cos3x)/d(3x) × d(3x)/dx = d(cosz)/dz × 3 (where z=3x) = -3 sinz = -3 sin3x.

### Q2: What is derivative formula of cos3x?

Answer: The derivative of cos3x formula is given by d(cos3x)/dx = -3sin3x.