Differential Review

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Implicit Differentiation

Derivatives aren't just for those functions whose formulas we already know. We can meaningfully ask for the rate of change of anything.

In particular, if a certain equation holds for all values of $x$, then the derivative of the left hand side must equal the derivative of the right hand side. If the expressions involve $y$, then the derivatives will involve $\displaystyle y'=\frac{dy}{dx}$ by the chain rule. Then solve for $\displaystyle \frac{dy}{dx}$ by putting all the terms that include $\displaystyle \frac{dy}{dx}$ on one side of the equation, and all the terms that don't on the other side.

Example: Find $y'$ where $$x^2y + y^3 =\sin(x).$$
DO: Try to work this problem before reading the solution.

Solution: Differentiating on both sides, using the product rule on the first term and the chain rule on the second, we get $$2xy + x^2 y' + 3y^2y' = \cos(x).$$ After grouping, we have $$\left(x^2+3y^2\right)y' = \cos(x)-2xy,$$ or $$\displaystyle{y' = \frac{\cos(x)-2xy}{x^2+3y^2}}.$$Note that the answer is typically an expression involving both $x$ and $y$. Occasionally we can simplify this into something that just involves $x$, but usually we can't.

There are 2 main uses of implicit differentiation:

The first is to get information about curves where $x$ and $y$ are related in a way that's more complicated than just the function $y=f(x)$. Many curves are not functions (such as a spiral, or a circle, etc.), but we still might want to know the rate of change of the curve at a point. At each point $(x,y)$ on the curve, we can figure out the derivative, plot the tangent line, and estimate what the curve is doing nearby.

The second use is to compute the derivatives of inverse functions. If $y = f^{-1}(x)$, then $x = f(y)$, so $1 = f'(y) y'$, so $y' = 1/f'(y)$. Often that can be expressed in terms of $x$. We used this to compute the inverse trig functions, which we review here:

Example: Compute the derivative of $\tan^{-1}(x)$.

DO: Try to work this problem before reading the solution.

Solution: We write \begin{eqnarray*}y &=& \tan^{-1}(x) \cr\cr x &=& \tan(y) \cr \cr1 &=& \sec^2(y)\, y' \cr\cr y' &=& \frac{1}{\sec^2(y)} \cr\cr y' &=& \frac{1}{1+\tan^2(y)}\cr \cr y' &=& \frac{1}{1+x^2}\end{eqnarray*}The derivatives of $\ln(x)$, $\sin^{-1}(x)$ and $\sec^{-1}(x)$ can be derived similarly.

Another way to say this is that, since $x=f(y)$, $\displaystyle \frac{dx}{dy} = f'(y)$. However, $$\displaystyle \frac{dy}{dx} = \frac{\quad 1\quad }{\frac{dx}{dy}} = \frac{1}{f'(y)}.$$ In other words, $\displaystyle \frac{dx}{dy}$ is the reciprocal of $\displaystyle \frac{dy}{dx}$.

The Six Pillars of Calculus

Trigonometry Review

Exponential Functions

Logarithms and Inverse functions

Intro to Limits

Limit Laws and Computations

Continuity and the Intermediate Value Theorem

Limits at Infinity

Rates of Change

The Derivative Function

Basic Differentiation Rules

Product and Quotient Rules

Derivatives of Trig Functions

The Chain Rule

Implicit Differentiation

Derivatives of Logs

Derivatives in Science

Related Rates

Linear Approximation and Differentials

Differentiation Review

Absolute and Local Extrema

The Mean Value and other Theorems

$f$ vs. $f'$

Indeterminate Forms and L'Hospital's Rule

Optimization

Newton's Method

Anti-derivatives

The Area under a curve

Definite Integrals

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Implicit Differentiation