Question

Find the derivative from the first principle root cos 3x

Answer 1

d/dx rootCos3x
=1/2root(Cos3x)d/dx(Cos3x)
=-3sinn3x/2rootcos3x

Answer 2

So,The derivative is  $\frac{-3sin3x}{2\sqrt{cos3x}}$

Step-by-step explanation:

Given,

Find the derivative from the first principle of $\sqrt{cos3x}$

Differentiate above equation with respect to $x$,

∴$\frac{d}{dx} \sqrt{cos3x}$

$=\frac{d}{dx} cos^{\frac{1}{2} } 3x$

$=\frac{1}{2\sqrt{cos3x}} \frac{d}{dx}( cos3x)$  (∵$\frac{d}{dx} \sqrt{x}=\frac{1}{\sqrt{x} } \frac{d}{dx}(x)$)

$=\frac{1}{2\sqrt{cos3x}} -sin3x\frac{d}{dx}(3x)$ (∵ $\frac{d}{dx}cosx= -sinx$ )

$=\frac{-sin3x}{2\sqrt{cos3x}}3$ (∵$\frac{d}{dx}(3x)=3$ )

$=\frac{-3sin3x}{2\sqrt{cos3x}}$

∴ The derivative is  $\frac{-3sin3x}{2\sqrt{cos3x}}$