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:

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

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

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

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

= lim_h→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.

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

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

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

= 3cos3x × 0 – sin3x × 3 by the limit formulas lim_x→0 $\dfrac{\cos x -1}{x}$ =0 and lim_x→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.

Have You Read These 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 2^x: The derivative of 2^x is 2^xln2.

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