Derivatives Of Basic Elementary Functions

d/dx(g(x)) = d/dx(2√x) + d/dx(e^(3x)) + d/dx(ln(x^2)) = x^(-1/2) + 3e^(3x) + 2x^(-1)

The velocity v(t) of a particle is the derivative of its position function
s(t). Therefore, to find the velocity of the particle at time t=4, we need to evaluate v(4)
Given the position function s(t) = t^3-6t^2+9t+5.
The velocity function v(t) by taking the derivative with respect to time t:
v(t) = ds/dt = 3t^2 – 12t + 9
v(4) = 3.16 – 12. 4 + 9 = 9

Given the temperature function
T(t)=100e ^(−0.05t)
The rate of change of temperature is given by the derivative
dT/dt
dT/dt = 100 . -0.5 e^(−0.05t)
dT/dt (at t=10) = -5 e^(−0.05 * 10)​
dT/dt (at t=10) = -5 e^(−0.5)​ = 3.03

Using quotient rule,
d/dx(u/v) = (vu’ – uv’)/v^2
g'(x) = cos(x) . (1+cos(x)) – sin(x) . (-sin(x))/(1+cos(x))^2
= cos(x) + cos^2(x) +sin^2(x) /(1+cos(x))^2
= cos(x)+1/(1+cos(x))^2 = 1/(1+cos(x))

The product rule states that if you have two functions u(x) and
v(x), then the derivative of their product is given by:
(uv)’ = u’v + uv’
Applying the product rule,
u(x)=5x^4, v(x) = e^(2x)
d/dx(5x^4 e^(2x)) = 20x^3 . e^(2x) + 5x^4 . 2e^(2x)
= 20x^3 e^(2x) + 10x^4 e^(2x)

d/dx(1/x^-n) = d/dx(x^n) = nx^(n-1)

The chain rule states that if you have a composite function
f(g(x)), then the derivative is given by
f'(g(x)) . g'(x)

Here, f(u) = u^3, g(x) = 2x – 1
p'(x) = 3(2x-1)^2 . g'(x)
= 6(2x-1)^2

To find the marginal revenue, take the the derivative of the revenue function R(x) with respect to x and then evaluate it at x=4.
R'(x) = 16 – 2x
at x=4, R'(x) = 16-8 = 8

g(x) = 2x+3−1/x
Now, applying the rules, we get
g'(x) = 2 – 1/x^2​

Using the differentiation rules, d/dx(x^n) = n.x^(n-1)
d/dx(k) = 0
We get
d/dx(f(x)) = f'(x) = 15x^2 – 4x + 4