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
yes
nod/dx 1/x=lnx
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 $\displaystyle x^{-1}$
yes, but you forgot + Cint. e^-x= -1e^-x
the form in which you wrote the answer makes me think you got it by coincidence though
it seems it was coincidenceint. e^-2x= -2e^-2x
no, you took the derivative here. you want the integral.
$\displaystyle \int e^{kx}~dx = \frac 1ke^{kx} + C$ for $\displaystyle k \ne 0$ a constant
no, here you took the integral, not the derivatived/dx e^-2x= -1/2e^-2x
nope. see my comment two problems upint. e^2x=2e^2x
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
Hello, chocolatelover!
Could you please correct the following?
$\displaystyle \frac{d}{dx}(\ln x) \:=\:\frac{1}{x}$ . . . . Right!
$\displaystyle \frac{d}{dx}\left(\frac{1}{x}\right) \;=\;\frac{d}{dx}\left(x^{-1}\right) \;=\;-x^2 \;=\;-\frac{1}{x^2}$$\displaystyle \frac{d}{dx}\left(\frac{1}{x}\right) \:=\:\ln x$ . . . . no
$\displaystyle \int e^{-x}dx \:=\: -e^{-x} + C$ . . . . Yes!
Let $\displaystyle u \:=\:-2x\quad\Rightarrow\quad du \:=\:-2\,dx\quad\Rightarrow\quad dx \:=\:-\frac{du}{2}$$\displaystyle \int e^{-2x}dx \:=\: -2e^{-2x}+ C$ . . . . no
Substitute: .$\displaystyle \int e^u\left(-\frac{du}{2}\right) \;=\;-\frac{1}{2}\int e^u\,du \:=\:-\frac{1}{2}\,e^u + C \;=\;-\frac{1}{2}\,e^{-2x}+C$
$\displaystyle \frac{d}{dx}\left(e^{-2x}\right) \:=\:-\frac{1}{2}\,e^{-2x} +C$ . . . . Correct!
Let $\displaystyle u = 2x \quad\Rightarrow\quad du = 2\,dx \quad\Rightarrow\quad dx = \frac{du}{2}$$\displaystyle \int e^{2x}dx \:=\:2e^{2x} + C$ . . . . no
Substitute: .$\displaystyle \int e^u\left(\frac{du}{2}\right) \;=\;\frac{1}{2}\int e^u\,du \;=\;\frac{1}{2}\,e^u + C \;=\;\frac{1}{2}\,e^{2x} + C$