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

2 Answers
Adrian D.
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#

Jim H
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#

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