Question

how to solve derivative of sin3x using first principle?

Answer 1

Answer:

Step-by-step explanation:

Is this right.....⤵⤵

f'(x) = lim_(x->0) [ f(x+h) - f(x)]/h

so f'(x) = lim_(x->0) [sin 3(x+h) - sin(3x)]/h

= lim_(x->0)[2 sin ((3x+3h-3x)/2) cos((3x+3h+3x)/2)]/h

=lim_(x->0)[2 sin ((3h)/2) cos ((6x+h)/2)]/h

= lim_(x->0) 2 (sin((3h)/2))/(((3h)/2)*(2/3)) cos ((6x+h)/2)

= 3cos 3x

Answer 2

The derivative of sin3x using first principle is$3\cos 3x.$

Step-by-step explanation:

Let $y=\sin 3x$

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

Then, $y+\delta y=\sin 3(x+\delta x)$

⇒ $\delta y=\sin (3x+3\delta x)-\sin 3x$

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

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

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

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

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

=$3\cos 3x$

Hence, the derivative of sin3x using first principle is$3\cos 3x.$