Answer:
(x-2)^2 + (y - 5)^2 = 16
Step-by-step explanation:
Given the standard form of an equation of a circle
(x-h)^2+(y-k)^2=r^2
If (h, k) = (5,2) and r = 4
Substitute
(x-2)^2 + (y - 5)^2 = 4^2
(x-2)^2 + (y - 5)^2 = 16
Hence the required equation is (x-2)^2 + (y - 5)^2 = 16
To find the derivative of this function, there is a property that we should know called the Constant Multiple Rule, which says:
![\dfrac{d}{dx}[cf(x)] = cf'(x)](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdx%7D%5Bcf%28x%29%5D%20%3D%20cf%27%28x%29)
(where 
is a constant)
Remember that the derivative of 
is 
. However, you may notice that we are finding the derivative of 
, not 
. So, we are going to have to use the chain rule. To complete the chain rule for the derivative of a trigonometric function (in layman's terms) is basically the following: First, complete the derivative of the trig function as you would if what was inside the trig function is 
. Then, take the derivative of what's inside of the trig function and multiply it by what you found in the first step.
Let's apply that to our problem. Right now, I am not going to worry about the 
at the front of the equation, since we can just multiply it back in at the end of our problem. So, let's examine 
. We see that what's inside the trig function is 
, which has a derivative of 2. Thus, let's first find the derivative of as if was just and then multiply it by 2.
The derivative of would first be 
. Multiplying it by 2, we get our derivative of 
. However, don't forget to multiply it by the that we removed near the beginning. This gives us our final derivative of .
Remember that we now have to find the derivative at the given point. To do this, simply "plug in" the point into the derivative using the x-coordinate. This is shown below:
![-\cot[2(\dfrac{\pi}{4})]\csc[2(\dfrac{\pi}{4})]](https://tex.z-dn.net/?f=-%5Ccot%5B2%28%5Cdfrac%7B%5Cpi%7D%7B4%7D%29%5D%5Ccsc%5B2%28%5Cdfrac%7B%5Cpi%7D%7B4%7D%29%5D)
Our final answer is 0.
Whatever your choices are, just substitute the x and y values to the equations. If they satisfy the inequality, then they are the solutions. For example, point (7, 1) satisfies the inequalities.
y > -2x
1 > -2(7)
1 > -14 (true)
<span>3y < x – 2
3(1) < 7 - 2
3 < 5 (true)
I hope I was able to help you. Have a good day.</span>
