Answer:
![\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B-2%7D%7Bx%20%5Cln%20%2810%29%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
General Formulas and Concepts:
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Addition/Subtraction]: ![\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%20%2B%20g%28x%29%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%5D%20%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bg%28x%29%5D)
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Quotient Rule]: ![\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5B%5Cfrac%7Bf%28x%29%7D%7Bg%28x%29%7D%20%5D%3D%5Cfrac%7Bg%28x%29f%27%28x%29-g%27%28x%29f%28x%29%7D%7Bg%5E2%28x%29%7D)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify.</em>
![\displaystyle y = \frac{\log (x)}{\log (x) - 2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20%5Cfrac%7B%5Clog%20%28x%29%7D%7B%5Clog%20%28x%29%20-%202%7D)
<u>Step 2: Differentiate</u>
- [Function] Derivative Rule [Quotient Rule]:
![\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x) - 2]'[\log (x)]}{[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B%5B%5Clog%20%28x%29%20-%202%5D%5B%5Clog%20%28x%29%5D%27%20-%20%5B%5Clog%20%28x%29%20-%202%5D%27%5B%5Clog%20%28x%29%5D%7D%7B%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
- Rewrite [Derivative Rule - Addition/Subtraction]:
![\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x)' - 2'][\log (x)]}{[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B%5B%5Clog%20%28x%29%20-%202%5D%5B%5Clog%20%28x%29%5D%27%20-%20%5B%5Clog%20%28x%29%27%20-%202%27%5D%5B%5Clog%20%28x%29%5D%7D%7B%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
- Logarithmic Differentiation:
![\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - [\frac{1}{\ln (10)x} - 2'][\log (x)]}{[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B%5B%5Clog%20%28x%29%20-%202%5D%5Cfrac%7B1%7D%7B%5Cln%20%2810%29x%7D%20-%20%5B%5Cfrac%7B1%7D%7B%5Cln%20%2810%29x%7D%20-%202%27%5D%5B%5Clog%20%28x%29%5D%7D%7B%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
- Derivative Rule [Basic Power Rule]:
![\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - \frac{1}{\ln (10)x}[\log (x)]}{[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B%5B%5Clog%20%28x%29%20-%202%5D%5Cfrac%7B1%7D%7B%5Cln%20%2810%29x%7D%20-%20%5Cfrac%7B1%7D%7B%5Cln%20%2810%29x%7D%5B%5Clog%20%28x%29%5D%7D%7B%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
- Simplify:
![\displaystyle y' = \frac{\frac{\log (x) - 2}{\ln (10)x} - \frac{\log (x)}{\ln (10)x}}{[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Clog%20%28x%29%20-%202%7D%7B%5Cln%20%2810%29x%7D%20-%20%5Cfrac%7B%5Clog%20%28x%29%7D%7B%5Cln%20%2810%29x%7D%7D%7B%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
- Simplify:
![\displaystyle y' = \frac{\frac{-2}{\ln (10)x}}{[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B%5Cfrac%7B-2%7D%7B%5Cln%20%2810%29x%7D%7D%7B%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
- Rewrite:
![\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B-2%7D%7Bx%20%5Cln%20%2810%29%5B%5Clog%20%28x%29%20-%202%5D%5E2%7D)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
10 1/4 - 3 1/2 = 6 3/4. So it’s 6 3/4 inches above ground. Now to double check: 6 3/4 + 3 1/2 = 10 1/4
line segment connecting the vertices of a hyperbola is called the <u>transverse axis</u> and the midpoint of the line segment is the <u>center</u> of the hyperbola.
What is transverse axis and center of hyperbola ?
The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints.
And The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes.
Learn more about the transverse axis and center of hyperbola here:
brainly.com/question/28049753
#SPJ4
If two pentagons are similar, then their sides are proportional. To find PT, we'll need to set up a proportion.
AE / PT = AB / PQ
---There are many ways to set up this proportion, it just depends on what side lengths you have and ensuring that they match up on both shapes.
6 / x = 5 / 12.5
5x = 75
x = 15
Length of PT = 15 cm
Option C
Hope this helps!