Implicit Differentiation

This paragraph introduces the concept of implicit differentiation, which is the first lesson in Unit 4. The instructor explains that while previously derivatives were taken of explicit functions, implicit differentiation deals with equations where x and y are intertwined, such as y^2 = x or e^y = √x. The goal is to find the derivative dy/dx for these types of equations, which requires understanding the chain rule in a deeper context. The instructor also mentions common implicitly defined curves like the unit circle and hyperbola, and sets the stage for practicing derivative rules and investigating the chain rule further.

The instructor delves into the specifics of taking derivatives using the chain rule, emphasizing the importance of recognizing when to apply it. Examples given include the derivative of f(x) squared and the derivative of 3x to the fourth with respect to t, highlighting the need to account for the derivative of the inner function. The paragraph reinforces the concept that the derivative of x with respect to x is 1, which is foundational in implicit differentiation. The instructor also provides practice problems for students to try on their own, encouraging them to apply the chain rule and product rule as necessary.

This section outlines the steps for finding the derivative of an implicitly defined curve, including taking the derivative of both sides of the equation with respect to x, moving all terms involving dy/dx to one side, factoring out dy/dx, and solving for it. The instructor demonstrates the process with a specific equation and encourages students to try it with other examples. The importance of this technique is underscored as a means to find the slope of tangent lines to curves that are not explicitly defined in terms of y or x.

The instructor discusses how to write the equation of a tangent line to a curve at a given point, using the previously found derivative dy/dx. An example is provided where the slope of the tangent line to a specific curve at a given point is calculated, and the equation of the tangent line is derived. The concept of horizontal and vertical tangents is introduced, explaining that horizontal tangents occur where the slope is zero and vertical tangents occur where the slope is undefined due to division by zero.

The paragraph focuses on identifying the points of horizontal and vertical tangents on a curve by setting the numerator of dy/dx to zero for horizontal tangents and the denominator to zero for vertical tangents. The instructor works through an example, solving for y when the derivative equals zero and when the denominator of the derivative equals zero to find the specific points on the curve. This method is crucial for understanding the behavior of curves at certain points and for further analysis such as drawing tangent lines.

The instructor explains the process of finding the second derivative of a function to determine the concavity of a curve at a specific point, which is essential for understanding the curve's shape and identifying local maxima and minima. The second derivative test is highlighted, where a positive second derivative indicates concave up and a negative second derivative indicates concave down. An example is provided where the second derivative is calculated, and it is concluded that the curve has a local maximum at the given point due to the first derivative being zero and the second derivative being negative.

The final paragraph discusses the position of a tangent line relative to a curve, whether it lies above or below, which is determined by the concavity of the curve. The second derivative is used to ascertain this, with a negative second derivative indicating the tangent line lies above the curve (since the curve is concave down). The instructor also provides a quick example of finding the slope of a tangent line for an equation where sine of x equals e to the power of y, demonstrating implicit differentiation and the use of substitution to express the derivative solely in terms of x.