My Math Forum Showing limit equals derivative

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 September 26th, 2013, 07:37 AM #1 Newbie   Joined: Sep 2013 Posts: 1 Thanks: 0 Showing limit equals derivative This is giving me so much trouble and I have no idea why. I just have to show that f'(x) = ( f(x + e) - f(x - e) ) / 2e for x = 5 and e = .001 The actual function is f(x) = e^(sinx) + xln(3 + cosx) and I'm almost positive the derivative is f'(x) = cos(x)e^(sinx) + ln(3 + cosx) - xsinx/(3 + cosx) When I evaluate the derivative at 5 I get 2.363208918 but when I evaluate the function with the small e I get 1.402409569 I just can't figure out why these numbers don't match up. Thanks.
 September 26th, 2013, 10:45 AM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 633 Thanks: 94 Math Focus: Electrical Engineering Applications Re: Showing limit equals derivative Hi jpanderson8, and welcome to the forums. I think that the derivative is correct, however, I think you are making the calculation using degrees. If you use radians, you should get 2.75783 (for both calculations). Also, I know that it is difficult without using $\ \LaTeX \$ but just to make sure let me restate: $f'(x)=\frac{f(x+\epsilon)-f(x- \epsilon)}{2\epsilon} \qquad \text{with} \ \epsilon=0.001$ $\text{this makes} \ \epsilon \ \text{distinct from} \ e, \ \text{the base of the natural logarithm}$ .

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