U3 L1 1 Notes Implicit m

Summary

Dr. Zak introduces implicit differentiation, building on the basic derivative rules and chain rule from the previous unit. The central idea is to differentiate both sides of an equation with respect to a chosen variable—even when the variables are mixed together—and include a factor such as (\frac{dy}{dx}) whenever differentiating a (y)-term with respect to (x).

Main Concepts

• Derivative notation specifies what is changing relative to what. For example, (\frac{dy}{dx}) is the derivative of (y) with respect to (x), while (\frac{ds}{dt}) is the derivative of (s) with respect to (t).
• Explicit equations isolate one variable, such as (y=f(x)). Implicit equations have variables mixed together. Solving an implicit equation for one variable may be difficult or introduce domain restrictions, so differentiating it directly is useful.
• Use the chain rule for dependent variables. For example, differentiating (y^3) with respect to (x) gives (3y^2\frac{dy}{dx}). More generally, differentiate the outside function and multiply by the derivative of the inside variable.
• Use the product rule when multiplying expressions that contain variables. When differentiating with respect to (x), remember that the derivative of (x) is (1), while the derivative of (y) is (\frac{dy}{dx}).
• The derivative depends on the variable named in the instruction. For example, differentiating (y^3) with respect to (t) gives (3y^2\frac{dy}{dt}), not (3y^2\frac{dy}{dx}).

General Procedure for Implicit Differentiation

1. Identify the variable with respect to which you are differentiating.
2. Differentiate both sides of the equation with respect to that variable.
3. Apply the chain rule to every term involving a dependent variable. For example, differentiating (y) with respect to (x) introduces (\frac{dy}{dx}).
4. Apply the product rule wherever two expressions containing variables are multiplied.
5. Collect all terms containing the desired derivative on one side.
6. Factor out the derivative and solve for it.
7. If asked for the derivative at a point, substitute the point’s coordinates after finding the derivative.

Examples and Applications

• One example differentiates an equation involving (x^2), (xy), and (y^3). After applying the product and chain rules, the terms containing (\frac{dy}{dx}) are collected and factored. At ((0,1)), the resulting derivative is (\frac14).
• Another example finds points where the slope is (-3). The derivative is set equal to (-3), giving a relationship between (x) and (y). Substituting that relationship into the original equation identifies the corresponding points.
• The lesson also applies implicit differentiation to a trigonometric equation. After differentiating a cosecant expression, the instructor rearranges the result and rewrites reciprocal trigonometric functions using familiar identities.
• To find a tangent line, the instructor first rewrites an expression with a fractional exponent as an equivalent equation that is easier to differentiate. The derivative gives the tangent-line slope, and the point-slope formula completes the line.

Practice and Homework

Implicit differentiation takes practice. Students are asked to work selected problems, including problems with multiple product rules, check their answers with CalcChat, and attempt a “rock problem” assigned for homework.

Speakers and Sources

• Dr. Zak — instructor and speaker
• CalcChat — resource mentioned for checking answers