Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

$\sqrt{x+y}+\sqrt{x-y}=\frac{x^{2}}{y^{3}}+\frac{31}{27} .$ Determine the equation of the tangent line to the curve at $(5 / 2,3 / 2)$.

$$y=\frac{79}{373} x+\frac{362}{373}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Missouri State University

University of Michigan - Ann Arbor

University of Nottingham

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

03:13

Find an equation of the ta…

01:33

Find the equation of the t…

01:20

02:23

01:59

03:21

01:08

for this problem. We've been given an equation, and our goal is to find the equation of the tangent line to the curve at a given point and are given point is five halves, three halfs. So what do we need in order to write the equation of a line? Well, if we had a slope in an intercept, we could use the slope intercept form. We don't have either of those. What we do have is a point, so we can use the point slope form, which is why minus y one equals M times X minus X one. We've already been given our point. All we need is our slope. And fortunately for us, we know how to find the slope. The slope is the derivative. So we will take our equation, find the derivative plug that into our line equation along with our point and get our tangent line. Now, one thing to bear in mind is we're gonna be using implicit differentiation here. This would be a very difficult equation to solve. Just for why So we need to remember that why is a function of X so if I'm taking the derivative of something with why, for example, we have a white cube will be looking at in a minute. When I take that derivative, we get three y squared. But I have to multiply by D Y dx because of the chain rule. You know, why is it just a variable? It's a function. So I have to recognize that it's a function and take its derivative as well. With respect to X, anytime we come across a y taking the derivative, we're gonna attack on that D Y DX because of the chain rule. Okay, let's go back to our original equation here. Okay? One thing, actually. One other quick reminder before we do that when we're taking the derivative of a square root, remember that we can write that square root as a fractional exponents X to the one half. So when I take the derivative that one half comes down. So I have a two in the denominator. I subtract one from the exponents. So I have X to the negative one half. Well, that's the equivalent of putting that X to the one half of the square root in the denominator. And then if I have, if X is any kind of function. I would put that derivative on the top. But when we're going to do our square roots is just helpful to remember what those derivatives look like because we have two of them here. So let's take the derivative of our first square, Rich. Well, that's gonna have a two and the square root in the denominator numerator. I'm going to take the derivative of what's under the radical one Plus Now, the derivative of why is going to be one de y de accepts that change. While we were talking about same thing with our second radical two times the square root of X minus Why goes on the denominator and my numerator is one minus again D Y d x we need to put in there. Okay, now I have a quotient x squared over white cute. We need to use the quotient rule denominator times the derivative of the numerator plus the numerator times the derivative of the denominator. And again, I gotta Why? So it's d Y DX over the square of the denominator and the derivative of that Constance just going to be zero. Okay, I'm going to do a real quick simplification here. Just a little one on the right hand side. Every term on the top has a Y squared. So I can, um, simplify this. I'm gonna cancel out the y squared from the second piece. This white cube just gonna go down toe. Why? And the denominator is now going to be. Why? To the fourth. So just simplifying it a little bit now, if I just want the derivative, I would expand all of this out. Solve it for D Y d X. A little bit simpler is to say, I don't really want the derivative. I want the value of the derivative at a certain point. So let's substitute in our 0.0.5 halves and three halves for X and y and see if that simplifies the math at all. So this first term one plus D Y d X over well, the square root of X plus y x plus why is going to be four square root of that is two times the two is already here. My denominators for second Fraction one minus d Y d x. Now I'm subtracting x and y so that's gonna give me one square root of one is one times two is two. Okay. And what do I have over here? Well, I have two times x times. Why? So if I multiply X and Y times two, that's going to give me 15 halves. Now I have X squared times three. That's going to be 75 3rd. I'm sorry. That thirds on the bottom. Sorry, I've got to is on the bottom. So it's 75/4 times. D Y d X over. Why? To the fourth. Well, that's going to be 81 over 16, okay? And I'm just going to simplify this piece right here. Since I'm here, I don't wanna have a fraction over a fraction, so I'm gonna multiply by the reciprocal. That's the same as multiplying by 16/81. And I could get rid of that denominator. Okay, already, it's looking a little bit better. Not great, but better. Let's get rid of the denominators on the left hand side. I'm gonna multiply both sides by four. Okay, So when I dio over here on the left hand side, those they're going to cancel, So I get one. Plus d Y d x two is going to cancel, so it's gonna be multiplying by two. So it's gonna be plus two minus two D Y d X, And I know I could put these together. This is three minus D Y d x. That's going to be the right hand side. Okay. What about the left? Well, I could do a little bit of canceling. Okay. When I multiply these out, I'm just gonna multiply everything here. So this is going to be this piece right here. This is 64/81 Going to multiply by both of my terms here and I'm going to get and I'm going to just come right down to here 1 60 over 27th, minus 400 over 27 d Y d x. So, so far, you could see these problems had ah, little bit of calculus and a lot of algebra and a lot of arithmetic. Now we're going to solve for y DX all the d y DX terms goto one side. So this will give me 427th d Y d x minus D Y d x. The right hand side gets my constance 160 over 27 minus three. Yeah, Well, let's combine these 400 over 27 d Y DX minus one are going to get a common denominator. This is going to give me 373 27th equaling extend. Same thing. Other side common denominator combined them and I get 79/27. Last step. I'm gonna multiply both sides by the reciprocal of that coefficient. And that gives me de y DX equaling 79 over 373. Not a very pretty number, but that's our slope at for our particular equation at a given point. This is our slope. So now we're going to use our slope intercept form as a reminder. Here's our point. Five halves, three halves. So when I go to write my equation for my line, let's do great here. Remember, it's why minus why one equals M times X minus X one. So why minus three half That's R Y. Coordinate. 79 overs. 3. 73 is our slope X minus five halves. Hey, now, usually we put these into slope Intercept form has just kind of, ah accepted way of writing these in a nice, simple form. I want to get rid of my parentheses. So that gives me 395 over 746 and last step. I'm gonna add three halves to both sides. I will need I'm sorry to read that nine there. I will need a common denominator. And when I do that and combine my numbers, I end up with a minus. I'm sorry. The plus 362 over 373. So that is the equation off our tangent line.

View More Answers From This Book

Find Another Textbook

02:15

Consider the function given by$$f(x)=\left\{\begin{aligned}-2 x+1 &a…

01:10

Show that if two functions $f$ and $g$ have the same derivative on the same …

01:11

Use the first and second derivatives to sketch the graph of the given equati…

02:21

Find the equation of the tangent line to the curve at the given $x$ -value.<…

04:09

Sketch the graph of the function defined by the given equation.$$f(x)=\f…

01:48

If $f(x)=\sqrt{2 x-1},$ find $(a) f^{-1}(1)$,(b) $f^{-1}(3)$

01:04

Write the given expression in logarithmic format.$$2^{4}=16$$

04:36

$$5 s^{2}\left(v^{3}-1\right)=7 . \text { Find }(a) d s / d v;(b) d v / d s$…

01:25

Solve for $x$ in.$$9^{3 x+2}=27^{2+x}$$

06:34

Use linearization to approximate the given quantity. In each case determine …