Answer:

General Formulas and Concepts:
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: ![\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bf%28x%29g%28x%29%5D%3Df%27%28x%29g%28x%29%20%2B%20g%27%28x%29f%28x%29)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>

<u>Step 2: Differentiate</u>
- [Function] Derivative Rule [Product Rule]:
![\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B9x%5E%7B10%7D%5D%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Rewrite [Derivative Property - Multiplied Constant]:
![\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%209%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E%7B10%7D%5D%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Basic Power Rule:
![\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%2090x%5E9%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Arctrig Derivative:

Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Answer:
4Joules
Step-by-step explanation:
According to Hooke's law which states that extension of an elastic material is directly proportional to the applied force provide that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
K is the elastic constant
e is the extension
If a spring exerts a force of 6 N when stretched 3 m beyond its natural length, its elastic constant 'k'
can be gotten using k = f/e where
F = 6N, e = 3m
K = 6N/3m
K = 2N/m
Work done on an elastic string is calculated using 1/2ke².
If the spring is stretched 2 m beyond its natural length, the work done on the spring will be;
1/2× 2× (2)²
= 4Joules
The answer is D one third of AB.
Answer:
.
Step-by-step explanation:
2(3x^2+23x+14)
2(3x^2+21x+2x+14)
2(3x(x+7)+2(x+7)
2(x+7)(3x+2)
Answer:
a its the first one
Step-by-step explanation:
because i did the math and i have dont this question before