Thread: Some derivative questions

I could use some help on the following derivative questions.....



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Originally Posted by redraider717

I could use some help on the following derivative questions.....



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What help, specifically, do you need?

The rule that is being tested here is
$\displaystyle \frac{d}{dx}e^x = e^x$.

For example, the first one is just the product rule:
$\displaystyle f(x) = (x + 3)e^x$

$\displaystyle f^{\prime}(x) = 1 \cdot e^x + (x + 3)e^x = (x + 4)e^x$

-Dan

It's e^-x

I'm not sure what the derivative of e^-x is

I'm also not sure what to do with 2e^x

Originally Posted by redraider717

It's e^-x

I'm not sure what the derivative of e^-x is

in general, $\displaystyle \frac d{dx}e^{kx} = ke^{kx}$

here, k = -1

I'm also not sure what to do with 2e^x

recall that $\displaystyle \frac d{dx}cf(x) = c \frac d{dx}f(x)$ where $\displaystyle c$ is a constant.

thus, $\displaystyle \frac d{dx}2e^x = 2 \frac d{dx}e^x = 2e^x$

(basically this means, when deriving something with a constant coefficient, you can pretty much ignore the constant, find the derivative of the function, and then affix (multiply by) the constant again.

Also on 3/x+3 wouldn't this be the answer....


(3)'(x+3)-(x+3)'(3)/(x+3)^2

(0)(x+3)-(1)(0)(3)/(x+3)^2

0/(x+3)^2= undefined....

Originally Posted by redraider717

Also on 3/x+3 wouldn't this be the answer....


(3)'(x+3)-(x+3)'(3)/(x+3)^2

(0)(x+3)-(1)(0)(3)/(x+3)^2

0/(x+3)^2= undefined....

first of all $\displaystyle \frac 0{(x + 3)^2}$ is not undefined.

secondly, the derivative of x + 3 is 1, so you would end up with $\displaystyle \frac {-3}{(x + 3)^2}$

I have another simple question.

What's the derivative of

3lnx?

Originally Posted by redraider717

I have another simple question.

What's the derivative of

3lnx?

Originally Posted by Jhevon

... recall that $\displaystyle \frac d{dx}cf(x) = c \frac d{dx}f(x)$ where $\displaystyle c$ is a constant...

what do you think?

3(1/x)?

Originally Posted by redraider717

3(1/x)?

correct. or we could say 3/x, same thing, but it looks nicer

Also do you happen to use webwork or know how to type in

cubed root of 3

and is 1/e^x not 1/e^x?

Originally Posted by redraider717

Also do you happen to use webwork or know how to type in

cubed root of 3

i do not know how it is in webwork, but it is probably 3^(1/3) or something like that. i don't know if cubert(3) would work either. shouldn't there be some kind of tutorial for how to enter things into your webwork program?

and is 1/e^x not 1/e^x?

what?! you wrote the same thing twice, of course they are the same

1/(e^x) = e^(-x) if that makes it any easier for you

another common way to type e to some power is to type exp(some power)

so 1/(e^x) = e^(-x) = exp(-x)

Well the actual problem has e^-x in it but I knew that was 1/e^x and the derivative of e^x is e^x so I figured (1/e^x)' was 1/e^x

Originally Posted by redraider717

Well the actual problem has e^-x in it but I knew that was 1/e^x and the derivative of e^x is e^x so I figured (1/e^x)' was 1/e^x

no, that is wrong. if you changed it to 1/(e^x) you would have to use the quotient rule.

just use the rule i told you: $\displaystyle \frac d{dx}e^{kx} = ke^{kx}$

here, k = -1