Find the equation of the tangent line at the given point on each curve. x^2 y^2=81, (-1,9)

Name: Problem 19, Chapter 6: Calculus with Applications
Uploaded: 2020-03-15T04:44:42-08:00
Duration: 4 min 12 s
Channel: Shamshad Waris
Description: Find the equation of the tangent line at the given point on each curve. x^2 y^2=81, (-1,9)

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

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Key Concepts

Implicit Differentiation

Implicit differentiation is a method for finding the derivative of a function when the equation is given in a form where the dependent and independent variables are intermingled. This technique involves differentiating both sides of the equation with respect to the independent variable and solving for the derivative, treating the dependent variable as a function of the independent variable.

Tangent Line

The tangent line to a curve at a particular point is the straight line that touches the curve at that point and has the same instantaneous slope as the curve does there. It provides a linear approximation of the curve in the vicinity of that point.

Point-Slope Form

The point-slope form is a way to write the equation of a line when a point on the line and the slope are known. It is expressed as y - y? = m(x - x?), where (x?, y?) is a known point on the line and m represents the slope.

Derivatives as Slopes

The derivative of a function at a particular point gives the slope of the tangent line to the curve at that point. This concept is used to determine how steep the curve is at a given point and is central to finding the equation of the tangent line.

Transcript

00:01 To solve this problem, first we have to understand the equation of the tangent line.

00:06 Y is equal to m into x minus x1 plus y1.

00:12 X1 and y1 point on the line.

00:16 Okay and m is the slope of the tangent line.

00:20 We can compute the slope using the d -y by d x, in a derivative of y with respect to x.

00:27 So, using this formula, we can compute the equation of the tangently.

00:33 First, we have to compute the derivative of y with respect to x.

00:38 So we can take the derivative of this equation with respect to x.

00:43 So we can write here d by d x x square y square into d y by d x a t one.

00:54 Now here we can apply the product rule of the derivative.

00:58 As you know, the product rule of the derivative d by d x into y square plus y square into d by d x into x square.

01:12 As you know, the derivative of the constant term is equal to.

01:18 Now, we can solve this problem.

01:21 So we can write here x square.

01:23 As you know the derivative of y square with respect to x, we can write derivative of y into 2a, 2 .y.

01:30 Into d -y by d -y by d -a -x because here you have to apply the channel of the derivative because here is given y square that means the function of x with respect to x...

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