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
$\displaystyle \frac{dy}{dx}$.
