Use implicit differentiation to find an equation of the...

Key Concepts

Chain Rule

In the process of differentiating composite functions, the chain rule is essential. It is used when differentiating a function of y with respect to x, where y itself is a function of x. The rule states that the derivative of the composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function (dy/dx).

Product Rule

When differentiating an expression that is the product of two functions of x (or functions of x and y), the product rule is employed. This rule states that the derivative of a product is given by the derivative of the first function times the second plus the first function times the derivative of the second, and is crucial in implicit differentiation when such products appear.

Evaluating the Derivative at a Specific Point

After finding a general expression for the derivative dy/dx through implicit differentiation, it is common to substitute a particular (x, y) value to determine the numerical slope of the tangent line at that point. This step is key in constructing the precise equation of the tangent line.

Implicit Differentiation

This is a method used to differentiate equations where y is not isolated on one side of the equation but is instead given implicitly as a function of x. It allows one to take the derivative of both sides of an equation with respect to x, applying the necessary differentiation rules to terms involving y, and then solving for dy/dx.

Tangent Line Equation

The tangent line at a point on a curve provides the best linear approximation of the function near that point. Its equation is determined by its slope, given by the derivative at that point, and by passing through the specific point, often expressed in point-slope form.

Transcript

00:01 But this problem, you're supposed to use implicit differentiation to find the equation of the tangent line at the given point.

00:09 And so i think the first thing i would do with this equation is i would go ahead and multiply the y -squared again.

00:22 So the y -square times six would be six -y -squared and then minus x -y -squared is equal to x -qued.

00:33 And then when you go to use implicit differentiation, on your first term, you're going to take the derivative of 1 .4.

00:41 So 2 times 6 would be 12 y.

00:46 But since it's with respect to x, we need to do a y prime.

00:51 And then minus, you have to use the product rule here.

00:56 And so you're going to do the derivative of x, which would be 1, and you're going to keep the minus and the y squared.

01:03 And then you're going to do the derivative of a y, so y squared.

01:06 So you're going to do the y squared is 2y.

01:09 So you keep the minus x, and then you have, to y as a derivative of y squared.

01:18 But again, since you're doing this with respect to x, you need to do a y prime, and then derivative x cubed is 3x squared.

01:33 And then i would try to group your y prime together, and that way you can solve the equation for y prime.

01:57 I'm going to add my y squared to both sides.

02:03 I'll get rid of that.

02:09 And then i'm going to factor out of y prime, and then i'm going to divide both sides of the equation through by the 12 y minus 2 xy.

02:42 So you're going to get y prime that is equal to 3x squared plus y squared over 12 y minus 2 xy...

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