Find the tangents to the implicitly defined curve x^2 y+x y^2=6 at points for which x=1. Also, c

Find the tangents to the implicitly defined curve x^2 y+x...

Key Concepts

Second Derivative in Implicitly Defined Functions

The second derivative in the context of implicitly defined functions is obtained by differentiating the first derivative (dy/dx) with respect to x. This process often requires the use of both the product and chain rules and may involve substituting back the expression for dy/dx. The second derivative provides information about the concavity and the rate of change of the slope of the curve.

Tangent Line

The tangent line to a curve at a given point is a line that touches the curve at that point and has the same slope as the curve there. Calculating the tangent involves finding the first derivative (dy/dx) to determine the slope at the point, and then using the point-slope form of a line to write the equation of the tangent.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions. In implicit differentiation, because y is a function of x, differentiating any term involving y requires the application of the chain rule. This rule ensures that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

Product Rule

The product rule is used to differentiate the product of two functions. In the context of implicit differentiation, when the equation involves the product of x and y or their powers, the product rule helps differentiate these terms correctly. It states that the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x).

Implicit Differentiation

Implicit differentiation is a technique used when differentiating equations that define y implicitly as a function of x rather than explicitly. It involves differentiating both sides of an equation with respect to x while treating y as an implicit function of x, and then solving for dy/dx. This method is essential when dealing with equations where isolating y is difficult or impossible.

Transcript

00:02 Problem 11 here is an implicit differentiation problem where we'll actually have to differentiate implicitly twice.

00:08 We'll need to locate some points on the curve that we can use to find some slope values to write equations of tangent lines as asked in the problem.

00:18 So we'll start with this to implicitly define curve and differentiate implicitly by taking the derivative with respect to x of both sides of the equation.

00:29 So we need to use product rule here, 2xy plus x squared, d, y, then product rule again, the derivative of x is 1 with respect to x times the second piece plus the derivative of the second piece using chain rule here, incorporating the dydx times the first piece, and the derivative of a constant is always zero.

01:05 So we'll need to isolate our dydx terms so we can factor out that dydx and we'll want to move those terms without dydx to the other side.

01:15 So over here we'll have x squared dydx plus 2x y, xydx equals negative y squared minus 2xy we want to factor this out so we can work to isolate it and then ultimately divide both sides by it to get the ydx alone.

01:56 So now we've got what we can use to find the derivative of our, or sorry, to find the slope of this implicitly defined curve, the slope of the tangent line.

02:10 But you'll notice we've only been given an x coordinate.

02:13 So i need both an x and y coordinate to plug in here to the slope.

02:17 So in order to find what y coordinate i should use, i should go back to my original definition here.

02:26 And to have x squared y plus x y squared equal six and so i want to plug one into here and find the y values that occur when x equals 1 so i get y squared plus y equals 6 which if i subtract 6 from both sides i can solve by factoring so y plus 3 times y minus 2 equals 0 so i get the values y equals 3, y equals positive 2.

03:06 So i've got points 1 negative 3 and 1 2.

03:13 So now i want to take my derivative and evaluate it at those two points.

03:26 If i plug 1 in for x and negative 3 in for y.

04:07 So i get negative 9 plus 6 over 1 minus 6, negative 3 over negative 5.

04:20 Or three -fifths.

04:23 So one equation, if i use that slope and the point we used, y minus y plus three equals three -fifths x minus one, then evaluating the derivative at the second point, i'll get negative two squared minus two times one, 2 over 1 squared 2 means 1 times 1.

04:57 0.

04:57 Negative 4 minus 4 over 1 plus 4, negative 8 fits.

05:05 Making that equation of the tangent line, y minus 2 equals negative 8 5ths times x minus 1...

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