Question

solve by the First Principle
sin3x step.by step using the increment in y and increment in x​

Answer 1

Let y=\sin 3xy=sin3x

Let \delta yδy be an increment in y, corresponding to an increment \delta xδx in x.

Then, y+\delta y=\sin 3(x+\delta x)y+δy=sin3(x+δx)

⇒ \delta y=\sin (3x+3\delta x)-\sin 3xδy=sin(3x+3δx)−sin3x

⇒\dfrac{\delta y}{\delta x} =\dfrac{\sin (3x+3\delta x)-\sin 3x}{\delta x}

δx

δy

=

δx

sin(3x+3δx)−sin3x

⇒ \frac{dy}{dx} = \lim_{x \to 0} \dfrac{\delta y}{\delta x}

dx

dy

=lim

x→0

δx

δy

= \lim_{\delta x \to 0} \dfrac{\sin (3x+3\delta x)-\sin 3x}{\delta x}=lim

δx→0

δx

sin(3x+3δx)−sin3x

= \lim_{\delta x \to 0} \dfrac{2\cos (3x+\delta x)\sin \delta x}{\delta x}=lim

δx→0

δx

2cos(3x+δx)sinδx

=3 \lim_{\delta x \to 0} \dfrac{2\cos (3x+\delta x)\sin \delta x}{\delta x}=3lim

δx→0

δx

2cos(3x+δx)sinδx

=3\cos 3x3cos3x

Hence, the derivative of sin3x using first principle is3\cos 3x.3cos3x.