slope of the tangent line using limit

Question

the instantaneous rate of change of f at x = 2 using the limit definition of a derivative given [y = sqrt(x^2-1)]

Answer 1

This is a classic Calculus I problem involving limits, conjugates, and the definition of a derivative. Let's first look at the limit definition of the derivative.

Step 0: Limit Definition of the Derivative

f′(a) = limₓ→ₐ (f(x) − f(a)) / (x − a)

'f' is the function and 'a' is the specific 'x' value.

Step 1: Definitions



We are asked to find the instantaneous rate of change of y = √(x² − 1) at x = 2. This means that we should make the following definitions:

f(x) = √(x² − 1), a = 2

Step 2: Substitute Into the Limit Definition Equation

f′(2) = limₓ→₂ (√(x² − 1) − √(2² − 1)) / (x − 2)

f′(2) = limₓ→₂ (√(x² − 1) − √3) / (x − 2)

Note that if we were to try to evaluate the limit by substituting x = 2, we would get 0 / 0. This is indeterminate. We must use algebra now.

Step 3: Multiply the Numerator and the Denominator By The Conjugate

(√(x² − 1) − √3) / (x − 2) · (√(x² − 1) + √3) / (√(x² − 1) + √3) = ((x² − 1) − 3) / ((x − 2)(√(x² − 1) + √3))

Step 4: Simplify and Factor Only the Numerator

x² − 1 − 3 = x² − 4 = (x − 2)(x + 2)

Step 5: Cancel the Common Term

((x − 2)(x + 2)) / ((x − 2)(√(x² − 1) + √3)) = (x + 2) / (√(x² − 1) + √3)

Step 6: Evaluate the Limit

Now that the (x-2) terms have been canceled, we can finally evaluate the limit by substituting x = 2.

f′(2) = (2 + 2) / (√(2² − 1) + √3) = 4 / (√3 + √3) = 4 / (2√3) = 2 / √3

2 / √3, or 2√3 / 3, is the instantaneous rate of change of y = √(x² − 1) at x = 2