Who's good at calculus?

J

jambone

Member
My calc professor offered 20 points extra credit on the final to anyone that could differentiate a chain rule problem, using the derivative. (His example was xe^x). I consider myself pretty good at math, but after an hour of trying I just can't get it.

So could anyone show the steps to differentiate xe^x using the definition of the derivative (f(x+h)-f(x))/h
 
for a second i was an idiot and thought it was logarithmic differentiation and was like "damn, you have hard high school teachers"
 
I'm at lim h->o (xe^x+h)-(e^x)/h + e^x+h

The right side is what you want... But i don't know how to cancel your - e^x and h... ill keep trying.
 
Well since you have to do it by the definition the teacher is really just testing your algebra skills. I am assuming that you know what the definiton of the derivative is. If you do just solve the limit.
 
theres a grade 10 in my drama class thats doing college calculus and getting 97% he might be able to help u
 
This kids right. To do the derivative of xe^x you use the product rule which is the derivative of u times v plus the derivative of v times u. u is x and v is e^x. So the derivative of x is 1 so the first term is e^x then the derivative of e^x is just e^x. so the second term is xe^x. so you get e^x + xe^x. And this isn't a chain rule problem it's a product rule problem. You could probably find that using the definition of the derivative, but that's why more work and impossible to explain online
 
sorry i read it wrong.

alright, so you want to find (fg)', and this can be done by taking the two limits of f and g and combining them. so

f'(x) = lim f(x+h)g(x+h) - f(x)g(x+h)+f(x)+g(x+h)-f(x)g(x)

h->0 ---------------------------------------------------------

h

which simplifies to

f'(x) = lim [f(x+h)-f(x)]g(x+h) + f(x)[g(x+h) - g(x)]

h->0----------------------------------------------

h

which simplifies to something you can handle

f'(x) = lim f(x+h) - f(x) * g(x+h) + f(x) * g(x+h)-g(x)

h->0------------- -----------------

h h

and i trust you can figure that out by yourself.
 
isnt it xe^(x-1)

product rule is bring down the power then subtract one from the power on top

i dont see where u gotta use the chain rule there.....
 
ya i dont know what the big deal is 20 extra credit points for that wow i wish i had your teacher. Okay so product rule says:

x*(e^x)' + x' * (e^x), with ' meaning derivative

that gives

x*e^x + 1* (e^x), or

(1+x)e^x

someone else explained it right in this thread also, i forget who
 
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