Find the equation of the tangent line at the given point on each curve. y+(√(x))/(y)=3 ; (4,2)

Name: Problem 22, Chapter 14: Finite Mathematics and Calculus with Applications
Uploaded: 2020-05-21T20:47:57-08:00
Duration: 3 min 42 s
Channel: Amy Jiang
Description: Find the equation of the tangent line at the given point on each curve. y+(√(x))/(y)=3 ; (4,2)

step 1 Okay, so we're asked to find our equation to our tangent line. So we're going to need to find ou

Find the equation of the tangent line at the given point on...

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Key Concepts

Chain Rule

The chain rule is used for differentiating composite functions. When a term involving a function of a function appears during differentiation, the chain rule states that the derivative is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.

Evaluating the Derivative at a Specific Point

Once the derivative is computed through implicit differentiation, substituting the given point into the derivative expression provides the slope of the tangent line at that point. This slope is then used along with the point-slope form to formulate the equation of the tangent line.

Tangent Line

A tangent line to a curve at a given point is a straight line that just touches the curve at that point without crossing over it locally and possesses the same slope as the curve at that point. It provides a linear approximation of the curve near the point of tangency.

Implicit Differentiation

Implicit differentiation is a method used when a function is given in an implicit form, not explicitly solved for one variable in terms of the other. By differentiating both sides of the equation with respect to the independent variable and applying differentiation rules, one can solve for the derivative of the dependent variable.

Transcript

00:01 Okay, so we're asked to find our equation to our tangent line.

00:05 So we're going to need to find our slope.

00:07 So we'll take the derivative of this following function, and then we'll plug that into our point slope form to find our equation for our tangent line.

00:17 Okay, so let's start by taking our derivative.

00:25 Okay, so that should it be actually an x.

00:28 Okay, so we get dy over dx, and then we'll use our quotient rule here for this term.

00:35 So the derivative of the square root of x, that's, 1 half x to the power of negative 1 half times y minus the derivative of y.

00:47 So that's d .y over dx times the square root of x over y squared.

00:58 And that's equal to 2.

01:02 Okay.

01:03 So we'll multiply both sides by y squared just to get rid of our fraction.

01:10 So we get the following.

01:27 So we get y to the power of, or actually that's this itself since we have over y squared.

01:32 Okay, so this is equal to 0.

01:39 And now let's move this term to our right -hand side so we can factor out a d -y over the x.

01:47 So we get y squared plus, actually that's minus x to 5 1⁄2 is equal to 1 -half, x to the power of negative 1 -half y.

02:01 The negative of that, so i want to, since i want to subtract it to the other side...

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