derrivatives
- Thread starter pistol
- Start date
Hello, pistol!
Your problem makes little sense.
Can you state the original problem?
You want the derivative of \(\displaystyle f(x) \:=\:e^x\sin x\)
Use the Product Rule.
We are given: .\(\displaystyle f(x) \:=\:A\arctan x + B \) . (where \(\displaystyle x\) is negative?)
Its derivative is: .\(\displaystyle f'(x) \:=\:\dfrac{A}{1+x^2}\)
The second derivative is: .\(\displaystyle f''(x) \:=\:\dfrac{-2Ax}{(1+x^2)^2}\)
Your problem makes little sense.
Can you state the original problem?
You want the derivative of \(\displaystyle f(x) \:=\:e^x\sin x\)
Use the Product Rule.
We are given: .\(\displaystyle f(x) \:=\:A\arctan x + B \) . (where \(\displaystyle x\) is negative?)
Its derivative is: .\(\displaystyle f'(x) \:=\:\dfrac{A}{1+x^2}\)
The second derivative is: .\(\displaystyle f''(x) \:=\:\dfrac{-2Ax}{(1+x^2)^2}\)
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
(Your swapping the position of x in the inequalities is confusing. Better would be "x>= 0" and "x< 0".)can anyone help me with that:
f(x) = e^xsinx, 0<=x
Aarctgx + B, x<0 has a derrivative in point a=0. Find A and B. Is there a second derrivative in point a=0?
thanks
For x>= 0, use the product rule: f'= e^x sin(x)+ e^x cos(x) and f''= [e^x sin(x)+ e^x cos(x)]+ [e^x cos(x)- e^x sin(x)]= 2e^x cos(x).
For x< 0, f'= A/(x^2+ 1) and f''= -2Ax/(x^2+1)^2.
Strictly speaking, to find the derivative at x= 0, (there is no point in saying "a= 0" when there is no "a" in the formulas) we should look at the one sided limits: for x= 0, h< 0, [f(0+h)- f(0)]/h= e^hsin(h)/h= e^h (sin(h)/h)) and you should know that the limit of sin(h)/h, as h goes to 0, is 1 so that limit is e^0= 1. For x= 0, h> 0, [f(0+h)- f(0)]/h= [A arctan(h)+ B- A arctan(0)- B]/h= A arctan(h)/h and now, as h goes to 0 arctan(h)/h goes to 1 (I did not recall that off hand myself but drew a quick graph) so the limit of the difference quotient is A.
If this function is to be differentiable at x= 0, we must have those two limits the same: 1= A. B can be any number.
With A= 1, the second derivative for x< 0 is -2x/(x^2+ 1)^2.
The second derivative, if it exists, is the derivative of the first derivative so to find the second derivative at x= 0 we can look at the difference quotients using the first derivative rather than the derivative.
For h>= 0, f'(0+h)- f'(0)= (e^h sin(h)+ e^h cos(h)- 1)/h= e^h[ sin(h)/h- (cos(h(- 1)/h]. Now both sin(h)/h and (cos(h)- 1)/h go to 1 so the limit, as h goes to 0 from above, is 0. For h< 0, f(0+ h)= [A/(h^2+ 1)- A]/h= (A- Ah^2- A)(h(h^2+ 1))= Ah/(h^2+ 1) which goes to 0 as h goes to 0.
Last edited: