Answer:
y = -3x - 9
Step-by-step explanation:
Point-slope equation for line of slope -3 that passes through (-2,-3):
y+3 = -3(x+2)
Rearrange equation to slope-intercept form
y+3 = -3x - 6
y = -3x - 9
Answer:
![\frac{37}{8}](https://tex.z-dn.net/?f=%5Cfrac%7B37%7D%7B8%7D)
Step-by-step explanation:
1
---------> ![\frac{11}{8}](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B8%7D)
3
--------->
(multiply by
) -----------> ![\frac{26}{8}](https://tex.z-dn.net/?f=%5Cfrac%7B26%7D%7B8%7D)
![\frac{11}{8} +\frac{26}{8} =\frac{37}{8}](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B8%7D%20%2B%5Cfrac%7B26%7D%7B8%7D%20%3D%5Cfrac%7B37%7D%7B8%7D)
Answer:
.
Step-by-step explanation:
To find the derivative of the function
you must:
Step 1. Apply the Quotient Rule ![\left(\frac{f}{g}\right)'=\frac{f\:'\cdot g-g'\cdot f}{g^2}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bf%7D%7Bg%7D%5Cright%29%27%3D%5Cfrac%7Bf%5C%3A%27%5Ccdot%20g-g%27%5Ccdot%20f%7D%7Bg%5E2%7D)
![\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=\frac{\frac{d}{dx}\left(e^x+1\right)\left(e^x-1\right)-\frac{d}{dx}\left(e^x-1\right)\left(e^x+1\right)}{\left(e^x-1\right)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28%5Cfrac%7Be%5Ex%2B1%7D%7Be%5Ex-1%7D%5Cright%29%3D%5Cfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28e%5Ex%2B1%5Cright%29%5Cleft%28e%5Ex-1%5Cright%29-%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28e%5Ex-1%5Cright%29%5Cleft%28e%5Ex%2B1%5Cright%29%7D%7B%5Cleft%28e%5Ex-1%5Cright%29%5E2%7D)
![\frac{d}{dx}\left(e^x+1\right)=e^x\\\\\frac{d}{dx}\left(e^x-1\right)=e^x](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28e%5Ex%2B1%5Cright%29%3De%5Ex%5C%5C%5C%5C%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28e%5Ex-1%5Cright%29%3De%5Ex)
![=\frac{e^x\left(e^x-1\right)-e^x\left(e^x+1\right)}{\left(e^x-1\right)^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Be%5Ex%5Cleft%28e%5Ex-1%5Cright%29-e%5Ex%5Cleft%28e%5Ex%2B1%5Cright%29%7D%7B%5Cleft%28e%5Ex-1%5Cright%29%5E2%7D)
Step 2. Simplify
![\frac{{e^{2x}-e^x-e^{2x}-e^x}}{\left(e^x-1\right)^2} \\\\\frac{-2e^x}{\left(e^x-1\right)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%7Be%5E%7B2x%7D-e%5Ex-e%5E%7B2x%7D-e%5Ex%7D%7D%7B%5Cleft%28e%5Ex-1%5Cright%29%5E2%7D%20%5C%5C%5C%5C%5Cfrac%7B-2e%5Ex%7D%7B%5Cleft%28e%5Ex-1%5Cright%29%5E2%7D)
Therefore,
![\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=-\frac{2e^x}{\left(e^x-1\right)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28%5Cfrac%7Be%5Ex%2B1%7D%7Be%5Ex-1%7D%5Cright%29%3D-%5Cfrac%7B2e%5Ex%7D%7B%5Cleft%28e%5Ex-1%5Cright%29%5E2%7D)