Differentiate implicitly and then determine the equation of the tangent line to the curve define

step 1 The underlying theme in this problem is the tangent line. And to find the equation with tangent

Differentiate implicitly and then determine the equation of...

Key Concepts

Implicit Differentiation

Implicit differentiation is used to find the derivative of an equation in which y is not isolated. Instead of rewriting the equation explicitly as y = f(x), both sides of the equation are differentiated with respect to x, treating y as a function of x and applying the necessary differentiation rules when y appears.

Chain Rule

The chain rule is applied when differentiating composite functions, particularly when differentiating functions of y (which is itself a function of x). For example, when differentiating y^n or other composite expressions involving y, the derivative is taken with respect to y first and then multiplied by y?.

Product Rule

The product rule is required when differentiating terms that involve the product of two functions of x (or a function of x and a function of y). It states that the derivative of a product u(x)v(x) is u?(x)v(x) + u(x)v?(x), which is essential for correctly differentiating products such as x^2 ln y or x^2 y^3.

Evaluating the Derivative at a Point

Once the implicit differentiation has been performed, the derivative expression often contains both x and y. Evaluating the derivative at a given point allows you to find the numerical value of the slope of the tangent line at that specific point on the curve.

Equation of the Tangent Line

After finding the slope of the tangent line from the evaluated derivative, the equation of the tangent line is formulated using the point-slope form. This form is given by y - y0 = m(x - x0), where m is the slope and (x0, y0) is the given point, representing the best linear approximation to the curve at that point.

Transcript

00:01 The underlying theme in this problem is the tangent line.

00:05 And to find the equation with tangent line, you need the, i'll write down point first.

00:10 Because they give you the ordered pair and the problem is 2 -1.

00:15 So we already have half the information that we need.

00:18 What we have to find is a slope.

00:20 And we find the slope by doing the derivative, d -y, d -x, at the ordered pair, 2 -1.

00:27 So now we can look at the problem at y to the fourth.

00:33 Minus 2x squared natural log of y minus 8x plus 4x squared y cube this is going to get ugly plus y so starting with the derivative you know 4 y cube d y d x but if you notice the next term it has a product in here so it's the product rule but the derivative of x squared is 2x so make that 4x leave natural log of y alone now you leave the 2x squared alone and the derivative of natural log of y is 1 over y, d y d y d x equals on the right side the derivative of negative 8 x is negative 8 we have another product rule so that would be 8 x y cubed plus well 3 times 4 to be 12 x squared y squared d y d y d x and the derivative of y is d y d x it's one d y d x but i've run out of room so now because this is already ugly enough i didn't do a second line is i'm going to plug in one and for all my y's two and for all of my xes and well you'll notice is one cube to still one times four is four d y the x now i'm not going to worry about this x because i know natural log of one is zero so that's that's going away and then two squared as four times that two would be negative eight one divided by one as 1.

02:17 So we have 8, d, y, d, d, x there.

02:20 And on the right side, we have negative 8.

02:23 Let's see, 8 times 2 times 1, it would be 16.

02:29 Let's see 2 squared is 4 times 1, 12 times 4 is 48, dy, dx, plus d y, dx...

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