Well, so far so good (although it's the chain rule you're using, not the product rule)awww sorry this isnt a homework actually im just thinking of a problem my teacher would do. hehe sorry
i got
d/dx=(e^{eeex})*(d/dx e^{eex})*(d/dx e^{ex})*(d/dx e^{x})
i just cant get past that point i dont know how to do the product rule with 3 values.
dont i have to use the product rule for the 3 that i havent take the d/dx of?Well, so far so good (although it's the chain rule you're using, not the product rule)
Now try computing each of the derivatives in that product (use the chain rule again).... what are (d/dx e^{eex}), (d/dx e^{ex}) and (d/dx e^{x})?
No, since these aren't products -- they are function compositions, so you would need to use the chain rule. Think of it this way: if f(x) = e^{x}, then e^{ex} = f(f(x)).dont i have to use the product rule for the 3 that i havent take the d/dx of?
"(d/dx e^{eex}), (d/dx e^{ex}) and (d/dx e^{x})?
I forgot to put this in my previous post. It doesn't make any sense to start and equation with d/dx. It's very much like saying [itex]\sqrt[/itex] = 3. Each of these symbols represents an operation that is yet to be performed.awww sorry this isnt a homework actually im just thinking of a problem my teacher would do. hehe sorry
i got
d/dx=(e^{eeex})*(d/dx e^{eex})*(d/dx e^{ex})*(d/dx e^{x})
i just cant get past that point i dont know how to do the product rule with 3 values.
That's the wrong way to go, since it would make the problem more difficult for no purpose. The basic function in this problem is already "the" exponential function, e^{x}.could we differentiate it like a^x by so it owuld be lna*a^x