Definition of Derivative Question
lim as h approaches 0
square root of (16+h) - 4 / h
Lim (√(16+h) - 4)/h as h --> 0
= Lim (√(16+h) - 4)/h * (√(16+h) + 4)/(√(16+h + 4)
= Lim (16+h - 16) / (h(√(16+h) + 4)
= lim 1/(√(16+h + 4)
= 1/(√16 + 4) = 1/8
To find the derivative of a function, we need to take the limit as the change in the variable approaches zero. In this case, we want to find the derivative of the function f(x) = √(16+x).
The derivative of a function f(x) can be thought of as the instantaneous rate of change at a specific point x. It represents the slope of the tangent line to the graph of the function at that point.
To find the derivative of f(x), we'll use the definition of the derivative:
f'(x) = lim(h→0) [f(x+h) - f(x)] / h
In our case, f(x) = √(16+x), so we need to plug this into the definition:
f'(x) = lim(h→0) [√(16+(x+h)) - √(16+x)] / h
Now, let's simplify the expression. We'll start by rationalizing the numerator:
f'(x) = lim(h→0) [(√(16+(x+h)) - √(16+x)) * (√(16+(x+h)) + √(16+x))] / [h * (√(16+(x+h)) + √(16+x))]
Using the difference of squares formula (a² - b² = (a + b)(a - b)), we can simplify further:
f'(x) = lim(h→0) [(16+(x+h)) - (16+x)] / [h * (√(16+(x+h)) + √(16+x))]
f'(x) = lim(h→0) [h] / [h * (√(16+(x+h)) + √(16+x))]
The h in the numerator cancels out with the h in the denominator:
f'(x) = lim(h→0) 1 / (√(16+(x+h)) + √(16+x))
Plugging in h = 0, we get:
f'(x) = 1 / (√(16+x) + √(16+x))
Simplifying the expression, we have:
f'(x) = 1 / (2√(16+x))
So, the derivative of √(16+x) is 1 / (2√(16+x)).