need to determine f'(x) of f(x)= 5e^(x/5)

is it 5e^(x/5) * ln 5((5-x)/25)?

as f(x)= 5e^(x/5) then f'(x)=e^(x/5)

What happened to the first 5?differentiates to 1?

let u = x/5

5 is just a constant. You move it out of the way, then take the derivative of e^u

it cant be x and you dont have to do u substitution because when we differentiate 5e^(x/5) then we get 5e^(x/5) but now we also have to differentiate x/5 also therefore we would just bring that in chain rule and bring it infront of (1/5)5e^(x/5)as a result it will be only e^(x/5) by using chain rule

oh okay.. so thats e^(x/5)dx..

yes u got it !!!

I said let u = 1/5 to make it easier. Just trying to help :(

haha.. Hyginkz must`ve got it.. ^_^

hero its all koo but u just makes it worse so let it be out of it but it was a good suggestion one can do that also!!!

ok what does u signify, I see it in my book but have not been able to make the connection...

it is just a substitution to make problem easier in other words look fancy with no fractions involved

it's just a variable substitution to make expressions easier to derivate

ok

The whole time @Mimi_x3 says nothing

so the constant 5 gets derivitated to 1?

face palm ^

Lol, i was doing it..i was slow..

then e remains e?

sorry this is kicking my rear

There's a general rule to follow when taking the derivative of an expression with a constant affixed to it. Go and re-read that part of your text.

Use the formula.. \[\large \frac{d}{dx} e^{ax+b} => e^{ax+b}*a \] so... \[5e^{\frac{x}{5}} => 5e^{\frac{x}{5}} * \frac{1}{5} \]

derivative of x/5 is 1/5, multiplied by the constant 5 to = 1, and e^x/5 is all that's left

-7x^2 would be -14x right?

yes

OK, I think I get it.

now to find the slope when x = a for f'(x)

slope 1...

Thanks

i don't understand your question..derivative of x?

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