Derivative of sin3x using first principle

The derivative of sin 3x using first principles is; 3cos(3x)

We want to find the derivative of sin 3x using first principles.

Step 1;

f'(x) =

Step 2; Expand the bracket to get;

f'(x) =

Step 3: According to trigonometric identities, we know that;

sin A - sin B =

Applying that to the answer in step 2 gives;

f'(x) =

Step 4; Simplify the brackets to obtain;

f'(x) =

Step 5; Rewrite the denominator to get;

f'(x) =

Step 6; Input the limit of 0 for h to get;

f'(x) =

⇒ f'(x) =

0/0 = 1. Thus;

f'(x) = 3cos(3x)

Read more about differentiation from first principles at; link

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

i got 86 for this

hope this !

if you need an explanation just ask