Answer:
(x - 6)^2 + (y + 4)^2 = 4
Step-by-step explanation:
formula for a circle
( x - h )^2 + ( y - k )^2 = r^2
(h,k) is the center
r is the radius
(6,-4) is center
2 is radius
Substitute in numbers and watch the negatives.
The answer for the table is
x 4 6 8
y 0 2 4
10 feet is the one i hop i help
Answer:
![\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}](https://tex.z-dn.net/?f=%20%5Crm%20%5Cdisplaystyle%20y%27%20%3D%20%20%202%20%7Be%7D%5E%7B2x%7D%20%20%20%2B%20%20%20%20%5Cfrac%7B1%7D%7Bx%7D%20%20%7Be%7D%5E%7B2x%7D%20%20%2B%202%20%5Cln%28x%29%20%7Be%7D%5E%7B2x%7D%20)
Step-by-step explanation:
we would like to figure out the differential coefficient of ![e^{2x}(1+\ln(x))](https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2B%5Cln%28x%29%29)
remember that,
the differential coefficient of a function y is what is now called its derivative y', therefore let,
![\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%20%3D%20%20%7Be%7D%5E%7B2x%7D%20%20%5Ccdot%20%281%20%2B%20%20%20%5Cln%28x%29%20%29)
to do so distribute:
![\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%20%3D%20%20%7Be%7D%5E%7B2x%7D%20%20%2B%20%20%20%5Cln%28x%29%20%20%5Ccdot%20%20%7Be%7D%5E%7B2x%7D%20)
take derivative in both sides which yields:
![\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%27%20%3D%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20%28%20%7Be%7D%5E%7B2x%7D%20%20%2B%20%20%20%5Cln%28x%29%20%20%5Ccdot%20%20%7Be%7D%5E%7B2x%7D%20%29)
by sum derivation rule we acquire:
![\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}](https://tex.z-dn.net/?f=%20%5Crm%20%5Cdisplaystyle%20y%27%20%3D%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20%20%7Be%7D%5E%7B2x%7D%20%20%2B%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20%20%20%5Cln%28x%29%20%20%5Ccdot%20%20%7Be%7D%5E%7B2x%7D%20)
Part-A: differentiating $e^{2x}$
![\displaystyle \frac{d}{dx} {e}^{2x}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%20%7Be%7D%5E%7B2x%7D%20)
the rule of composite function derivation is given by:
![\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20f%28g%28x%29%29%20%3D%20%20%5Cfrac%7Bd%7D%7Bdg%7D%20f%28g%28x%29%29%20%5Ctimes%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20g%28x%29)
so let g(x) [2x] be u and transform it:
![\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdu%7D%20%20%7Be%7D%5E%7Bu%7D%20%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bdx%7D%202x)
differentiate:
![\displaystyle {e}^{u} \cdot 2](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%20%7Be%7D%5E%7Bu%7D%20%20%5Ccdot%202)
substitute back:
![\displaystyle \boxed{2{e}^{2x} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%20%20%5Cboxed%7B2%7Be%7D%5E%7B2x%7D%20%20%7D)
Part-B: differentiating ln(x)•e^2x
Product rule of differentiating is given by:
![\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20f%28x%29%20%5Ccdot%20g%28x%29%20%3D%20f%27%28x%29g%28x%29%20%2B%20f%28x%29g%27%28x%29)
let
substitute
![\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20%20%5Cln%28x%29%20%20%5Ccdot%20%20%7Be%7D%5E%7B2x%7D%20%20%3D%20%20%5Cfrac%7Bd%7D%7Bdx%7D%28%20%5Cln%28x%29%20%29%20%7Be%7D%5E%7B2x%7D%20%20%2B%20%20%5Cln%28x%29%20%5Cfrac%7Bd%7D%7Bdx%7D%20%20%7Be%7D%5E%7B2x%7D%20)
differentiate:
![\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20%20%5Cfrac%7Bd%7D%7Bdx%7D%20%20%5Cln%28x%29%20%20%5Ccdot%20%20%7Be%7D%5E%7B2x%7D%20%20%3D%20%20%20%5Cboxed%7B%5Cfrac%7B1%7D%7Bx%7D%20%7Be%7D%5E%7B2x%7D%20%20%2B%20%202%5Cln%28x%29%20%20%7Be%7D%5E%7B2x%7D%20%7D)
Final part:
substitute what we got:
![\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }](https://tex.z-dn.net/?f=%20%5Crm%20%5Cdisplaystyle%20y%27%20%3D%20%20%20%5Cboxed%7B2%20%7Be%7D%5E%7B2x%7D%20%20%20%2B%20%20%20%20%5Cfrac%7B1%7D%7Bx%7D%20%20%7Be%7D%5E%7B2x%7D%20%20%2B%202%20%5Cln%28x%29%20%7Be%7D%5E%7B2x%7D%20%7D)
and we're done!