How to find first derivative of f(x)=2 sin (3x) + x?
2 Answers
Explanation:
Differentiate each term:
Using the chain rules for the second term we have:
With:
Together we have:
We are asked to find the derivative of
Explanation:
We need to evaluate:
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.
I will assume that you can show that the second limit is
# =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:
# = lim_(hrarr0)(2sin(3(x+h))- 2sin3x)/h +lim_(hrarr0)((x+h)-x)/h#
# = 6cos(3x) +1#