Find the slope of the tangent to the curve y=1 / √(x) at the point where x=4 .

Name: Problem 32, Chapter 3: Calculus
Uploaded: 2019-05-04T20:52:23-08:00
Duration: 4 min 41 s
Channel: Karl Schaefer
Description: Find the slope of the tangent to the curve y=1 / √(x) at the point where x=4 .

Find the slope of the tangent to the curve y=1 / √(x) at the...

Key Concepts

Tangent Line

The tangent line to a curve at a specific point is the line that just touches the curve at that point and has the same slope as the curve. The slope of the tangent line, given by the derivative at that point, describes the instantaneous direction of the curve.

Power Rule

The power rule is a basic differentiation technique used for functions of the form x raised to a power. It states that the derivative of x^n is n*x^(n-1), and it simplifies finding the derivative of functions that can be expressed as a power of x.

Derivative

A derivative represents the instantaneous rate of change of a function with respect to its variable and is defined as the limit of the difference quotient. It is fundamental in understanding how a function behaves in an infinitesimally small neighborhood of a point.

Transcript

00:01 In this problem, we're asked to find the slope of the tangent line to the curb y equals 1 over the square root of x at the point where x is equal to 4.

00:09 So to do this, i find it kind of easier sometimes to rewrite this as more of a function notation, f of x equals 1 over square root of x at x equals 4.

00:18 Because now in order to find the slope, we know that the slope is given as a limit.

00:23 And i just like to have the f so that i can write f of x plus h minus f of x over h, which is a nice expression.

00:31 So now we just need to plug in our function.

00:34 So f of x plus h is 1 over the square root of x plus h minus 1 over the square root of x all over h.

00:43 But in this problem we're given what x is.

00:45 We're given x equals 4.

00:47 So we can just plug that in for x, oh, right here and here.

00:52 So 1 over the square root of 2 plus h, excuse me, 4 plus h, not 2 plus h, minus 1 over the square root of 4 all over h.

01:07 So this is a limit as h goes to 0 of 1 over the square root of 4 plus h minus 1 half, minus 1 half, all over h.

01:20 So now in order to solve this limit, we're going to need to combine the fractions in the numerator so that we can get it down to like a single expression.

01:28 This is a limit as h goes to 0 of 2 over 2 times the square root of 4 plus h.

01:35 So we're just going to put everything in the numerator with a common denominator, square root of 4 plus h over 2 times a square root of 4 plus h, all over h.

01:48 So once we combine this, this is the limit as h goes 0 now of 2 minus the square root of 4 plus h over 2 times a square root of 4 plus h, all over h.

02:02 Now let's combine these denominators, make this a little bit simpler.

02:07 This is limited as h goes 0 of 2 minus the square root.

02:09 Of 4 plus h divided by 2 times h times a square root of 4 plus h so all we've done here is we we've combined these two denominators by multiplying them together so 2 root 2 times a square of 4 plus h times h is 2h times square to 4 plus h so these two have been combined into this great this limit this limit still looks a little gnarly we're not entirely done yet whenever let's just just go go to a new page and rewrite our limit.

02:41 We need a little bit more space...

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