How to find first derivative of f(x)=2 sin (3x) + x?

2 Answers
Aug 7, 2015

f'(x)=6cos(3x)+1

Explanation:

Differentiate each term:

(d(x))/dx=1

Using the chain rules for the second term we have:

g(x)=h(k(x))=>g'(x)=k'(x)h'(k(x))

With:
h(u)=2sin(u)=>h'(u)=2cos(u)
k(x)=3x=>k'(x)=3
g(x)=2sin(3x)=>g'(x)=6cos(3x)

Together we have:
f'(x)=6cos(3x)+1

Aug 7, 2015

We are asked to find the derivative of f(x) = 2sin(3x)+x using the definition: f'(x) = lim_(hrarr0) ( f(x+h) - f(x))/(h).

Explanation:

We need to evaluate:

lim_(hrarr0) (overbrace([2sin(3(x+h))+(x+h)])^(f(x+h)) - overbrace([2sin(3x)+x])^f(x))/h.

This will be cumbersome. To make it look less complicated, let's split the expression into two simpler parts. We'll take the trigonometric part and the linear part separately.

lim_(hrarr0)(2sin(3(x+h))- 2sin3x)/h +lim_(hrarr0)((x+h)-x)/h

I will assume that you can show that the second limit is 1. The more challenging limit is the limit involving trigonometric functions.

lim_(hrarr0)(2sin(3(x+h))- 2sin3x)/h = 2lim_(hrarr0)(sin(3x+3h)- sin3x)/h

=2lim_(hrarr0)( overbrace((sin3xcos3h+cos3xsin3h))^sin(3x+3h)- sin3x)/h

=2lim_(hrarr0)(sin3xcos3x -sin3x +cos3xsin3x)/h

=2lim_(hrarr0)((sin3x(cos3h - 1))/h +(cos3xsin3h)/h)

=2lim_(hrarr0)(sin3x(cos3h - 1)/h +cos3x(sin3h)/h)

=2[lim_(hrarr0)sin3x lim_(hrarr0)(cos3h - 1)/h + lim_(hrarr0)cos3x lim_(hrarr0)(sin3h)/h]

=2[(lim_(hrarr0)sin3x) (3lim_(hrarr0)(cos3h - 1)/(3h)) + (lim_(hrarr0)cos3x) (3lim_(hrarr0)(sin3h)/(3h))]

=2[(sin3x) (3*0) + (cos3x) (3*1)]

= 2(3cos3x) = 6cos(3x)

So, when we put the two pieces together, we get:

f'(x) = lim_(hrarr0) ([2sin(3(x+h))+(x+h)] - [2sin(3x)+x])/h

= lim_(hrarr0)(2sin(3(x+h))- 2sin3x)/h +lim_(hrarr0)((x+h)-x)/h

= 6cos(3x) +1