Need to find the second derivative of:

[619snake]

f(x)= 1/125(e^5x)(5x-2)I think this is the first derivative but I ain't good at math and gives me some headaches I used the product rule.. but still I have doubts =(

f '(x)= 1/125(5)(e^5x)(5x-2) + (5)(1/125(e^5x)

Please help me!
I need to understand how to do this

 

[Raskolnikov]

Assuming you have f(x) = \frac{1}{125}e^{5x}(5x - 2), you need to apply the product rule.

Given f(x) = g(x)h(x),

f'(x) = g'(x)h(x) + g(x)h'(x).

And you'll need the product rule again to find f''(x). Also, I'd leave the constant 1/125 out in front while taking the derivatives until the end.

 

[619snake]

Thanks Raskolnikov! Now I only need to identify which rule I need to use in each case.

 

[Raskolnikov]

f '(x)= 1/125(5)(e^5x)(5x-2) + (5)(1/125(e^5x)

Yep, you got it right (though I think you're overloading on parentheses :P)! For the second derivative, you have to use the product rule again on the first term since it has both a factor of e^5x and a factor of (5x - 2). However, you don't need the product rule for the 2nd term since it's just e^5x.

Try to find f''(x). Just post whatever answer you get or where you're getting stuck, and I'll help you from there if you need.

 

[619snake]

I'm not sure about this... as I said.. I haven't mastered this rules and is getting really messed up

f"(x)= (\frac{1}{125})(25)e^{5x}(5x-2) + 5(\frac{1}{125}(5)e^{5x}) + (\frac{1}{125}(5)e^{5x})...

I don't know... think I got confused

 

[Raskolnikov]

Close! You forgot another factor of 5 on the last term. Don't worry, you'll get better with practice.

So it should simplify to:

f''(x) = \frac{1}{5}e^{5x}(5x - 2) + \frac{2}{5}e^{5x}.