Use long division if you have to.
The answer is 60
Answer:
it will be
Step-by-step explanation:
6/9 but a mix number i don't know
Answer:
A and B are correct, but A is most correct.
Step-by-step explanation:
just do Pythagorean theorem and see which numbers add up to each other. if you do them for all 4 answers, you see d and c do not equal to each other. b is almost correct, and a comes out exact.
By definition of the derivative,
![y'=\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^x}h](https://tex.z-dn.net/?f=y%27%3D%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B%28x%2Bh%29e%5E%7Bx%2Bh%7D-xe%5Ex%7Dh)
We can employ the standard manipulation for proving the product rule:
![\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^{x+h}+xe^{x+h}-xe^x}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B%28x%2Bh%29e%5E%7Bx%2Bh%7D-xe%5E%7Bx%2Bh%7D%2Bxe%5E%7Bx%2Bh%7D-xe%5Ex%7Dh)
![\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^{x+h}}h+xe^x\lim_{h\to0}\frac{e^h-1}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B%28x%2Bh%29e%5E%7Bx%2Bh%7D-xe%5E%7Bx%2Bh%7D%7Dh%2Bxe%5Ex%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Be%5Eh-1%7Dh)
![\displaystyle\lim_{h\to0}e^{x+h}\lim_{h\to0}\frac{(x+h)-x}h+xe^x\lim_{h\to0}\frac{e^h-1}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7De%5E%7Bx%2Bh%7D%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B%28x%2Bh%29-x%7Dh%2Bxe%5Ex%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Be%5Eh-1%7Dh)
![e^{x+0}\displaystyle\lim_{h\to0}\frac hh+xe^x\lim_{h\to0}\frac{e^h-1}h](https://tex.z-dn.net/?f=e%5E%7Bx%2B0%7D%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%20hh%2Bxe%5Ex%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Be%5Eh-1%7Dh)
![e^x+xe^x\displaystyle\lim_{h\to0}\frac{e^h-1}h](https://tex.z-dn.net/?f=e%5Ex%2Bxe%5Ex%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Be%5Eh-1%7Dh)
The remaining limit is pretty well-known and has a value of 1. We can derive it from the definition of
,
![e=\displaystyle\lim_{n\to0}(1+n)^{1/n}](https://tex.z-dn.net/?f=e%3D%5Cdisplaystyle%5Clim_%7Bn%5Cto0%7D%281%2Bn%29%5E%7B1%2Fn%7D)
In the limit above, we substitute
, so that
. As
, we have
:
![\displaystyle\lim_{h\to0}\frac{e^h-1}h=\lim_{\eta\to0}\frac\eta{\ln(\eta+1)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Be%5Eh-1%7Dh%3D%5Clim_%7B%5Ceta%5Cto0%7D%5Cfrac%5Ceta%7B%5Cln%28%5Ceta%2B1%29%7D)
![\displaystyle=\lim_{\eta\to0}\frac1{\frac1\eta\ln(\eta+1)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D%5Clim_%7B%5Ceta%5Cto0%7D%5Cfrac1%7B%5Cfrac1%5Ceta%5Cln%28%5Ceta%2B1%29%7D)
![\displaystyle=\frac1{\lim\limits_{\eta\to0}\ln(\eta+1)^{1/\eta}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D%5Cfrac1%7B%5Clim%5Climits_%7B%5Ceta%5Cto0%7D%5Cln%28%5Ceta%2B1%29%5E%7B1%2F%5Ceta%7D%7D)
![\displaystyle=\frac1{\ln\left(\lim\limits_{\eta\to0}(\eta+1)^{1/\eta}\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D%5Cfrac1%7B%5Cln%5Cleft%28%5Clim%5Climits_%7B%5Ceta%5Cto0%7D%28%5Ceta%2B1%29%5E%7B1%2F%5Ceta%7D%5Cright%29%7D)
![=\dfrac1{\ln e}=\dfrac11=1](https://tex.z-dn.net/?f=%3D%5Cdfrac1%7B%5Cln%20e%7D%3D%5Cdfrac11%3D1)
After all this, we've shown that
![(xe^x)'=e^x+xe^x=e^x(x+1)](https://tex.z-dn.net/?f=%28xe%5Ex%29%27%3De%5Ex%2Bxe%5Ex%3De%5Ex%28x%2B1%29)