1. ## integrals and derivatives

Hello everyone,

Could you please correct the following?

d/dx lnx=1/x

d/dx 1/x=lnx

int. e^-x= -1e^-x

int. e^-2x= -2e^-2x

d/dx e^-2x= -1/2e^-2x

int. e^2x=2e^2x

Thank you very much

2. Originally Posted by chocolatelover
Hello everyone,

Could you please correct the following?

d/dx lnx=1/x
yes

d/dx 1/x=lnx
no

didn't you basically reverse what you did in the first question here? that's not taking the derivative, that's taking the anti-derivative

hint: think of 1/x as $x^{-1}$

int. e^-x= -1e^-x
yes, but you forgot + C

the form in which you wrote the answer makes me think you got it by coincidence though

int. e^-2x= -2e^-2x
it seems it was coincidence

no, you took the derivative here. you want the integral.

$\int e^{kx}~dx = \frac 1ke^{kx} + C$ for $k \ne 0$ a constant

d/dx e^-2x= -1/2e^-2x
no, here you took the integral, not the derivative

int. e^2x=2e^2x
nope. see my comment two problems up

you have been doing problems like these for a while now. you should be able to do them easily. maybe you forgot all the rules over your summer break? it happens, don't worry about it. but fix it, and fix it quick. please review your derivative and integration rules

3. Hello, chocolatelover!

Could you please correct the following?

$\frac{d}{dx}(\ln x) \:=\:\frac{1}{x}$ . . . . Right!

$\frac{d}{dx}\left(\frac{1}{x}\right) \:=\:\ln x$ . . . . no
$\frac{d}{dx}\left(\frac{1}{x}\right) \;=\;\frac{d}{dx}\left(x^{-1}\right) \;=\;-x^2 \;=\;-\frac{1}{x^2}$

$\int e^{-x}dx \:=\: -e^{-x} + C$ . . . . Yes!

$\int e^{-2x}dx \:=\: -2e^{-2x}+ C$ . . . . no
Let $u \:=\:-2x\quad\Rightarrow\quad du \:=\:-2\,dx\quad\Rightarrow\quad dx \:=\:-\frac{du}{2}$

Substitute: . $\int e^u\left(-\frac{du}{2}\right) \;=\;-\frac{1}{2}\int e^u\,du \:=\:-\frac{1}{2}\,e^u + C \;=\;-\frac{1}{2}\,e^{-2x}+C$

$\frac{d}{dx}\left(e^{-2x}\right) \:=\:-\frac{1}{2}\,e^{-2x} +C$ . . . . Correct!

$\int e^{2x}dx \:=\:2e^{2x} + C$ . . . . no
Let $u = 2x \quad\Rightarrow\quad du = 2\,dx \quad\Rightarrow\quad dx = \frac{du}{2}$

Substitute: . $\int e^u\left(\frac{du}{2}\right) \;=\;\frac{1}{2}\int e^u\,du \;=\;\frac{1}{2}\,e^u + C \;=\;\frac{1}{2}\,e^{2x} + C$

4. Thank you very much

Regards