Find the instantaneous velocity at x=6 if
f(x)=radical 2x+4
f(x) = √(2x+4)
f'(x) = 1/√(2x+4)
now just plug in your numbers
power rule and chain rule
If y = u^n
and u is a function of x,
y' = n u^(n-1) * u'
To find the instantaneous velocity at x=6, we need to calculate the derivative of the function f(x) with respect to x, and then evaluate it at x=6.
Here's how you can do that step by step:
1. Start with the given function:
f(x) = √(2x + 4)
2. Find the derivative of f(x) with respect to x, using the power rule for differentiation. Recall that the derivative of √u is (1/2) * (du/dx) * u^(-1/2):
f'(x) = (1/2) * (d/dx) * (2x + 4)^(-1/2)
3. Apply the derivative to the expression inside the parentheses:
f'(x) = (1/2) * (d/dx) * (2x + 4)^(-1/2)
= (1/2) * (-1/2) * (2x + 4)^(-3/2) * (d/dx)(2x + 4)
4. Simplify the expression:
f'(x) = (1/2) * (-1/2) * (2x + 4)^(-3/2) * 2
= -1 / (2√(2x + 4)^3)
5. Now we have the derivative function f'(x). To find the instantaneous velocity at x=6, substitute x=6 into f'(x):
f'(6) = -1 / (2√(2(6) + 4)^3)
= -1 / (2√(12 + 4)^3)
= -1 / (2√(16)^3)
= -1 / (2√16^3)
= -1 / (32√16)
= -1 / (32 * 4)
= -1 / 128
Therefore, the instantaneous velocity at x=6 for the function f(x) = √(2x + 4) is -1/128.