Implicit Differentiation Implicit Differentiation
Introduction
We will now find derivatives for equations. These are not necessarily functions, as we are not assigning outputs. We are setting two combinations of variables equal to each other, and we are graphing the points which satisfy that equality.
An example is the unit circle,
y^2 + x^2 = 1
. Note that for each
x
, there are (almost always) two
y
values. It is not a function.
Despite that, the unit circle could have a tangent line, and thus it should have a derivative.
Implicit Differentiation of the Unit Circle
To find the derivative of the unit circle, we will differentiate both sides of the equation
y^2 + x^2 = 1
with respect to
x
.
\begin{align*}
\frac{d}{dx}y^2 + \frac{d}{dx}x^2 &= \frac{d}{dx}1 \\\\
2y \cdot \frac{dy}{dx} + 2x &= 0 \\\\
2y \cdot y' + 2x &= 0 \\\\
2y \cdot y' &= -2x \\\\
y' &= \frac{-2x}{2y} \\\\
y' &= \frac{-x}{y}
\end{align*}
Note that
y
is present in the derivative. This should be intuitive. In a function, if we knew the input value, we always knew exactly what point on the graph we were referencing. Now that we are using functions, we need both the
x
and the
y
value to know where we are on the graph.
Using this derivative, we can find the slope of the tangent line at any point on the unit circle.
At (0, 1)
, the slope is 0
.
At (\sqrt{2}, \sqrt{2})
, the slope is -1
.
At (1, 0)
, the slope is -\frac{1}{0}
, which does not exist. (Vertical Asymptote) At this point, the unit circle cannot define its change with respect to
x
, as
x
is not changing.
Basic Problems
xy=1
. Find
y'
.
\begin{align*}
xy&=1 \\\\
\frac{d}{dx}xy&=\frac{d}{dx}1 \\\\
y + xy'&=0 \\\\
xy'&=-y \\\\
y'&=-\frac{-y}{x}
\end{align*}
e^{x+y}=x-y
. Find
y'
.
\begin{align*}
e^{x+y}&=x-y \\\\
\frac{d}{dx}e^{x+y}&=\frac{d}{dx}x-y \\\\
e^{x+y}\frac{d}{dx}[x+y]&=1-y' \\\\
e^{x+y}[1+y']&=1-y' \\\\
e^{x+y}[1+y']+y'&=1 \\\\
e^{x+y}+ e^{x+y} \cdot y'+y'&=1 \\\\
y' \cdot [e^{x+y} + 1]&=1-e^{x+y} \\\\
y'&=\frac{1-e^{x+y}}{e^{x+y} + 1}
\end{align*}
Intermediate Problems
\sin(xy) + y^2 = \cos(x)
. Find
y'
.
\begin{align*}
\frac{d}{dx}[\sin(xy) + y^2] &= \frac{d}{dx}\cos(x) \\\\
\cos(xy)\frac{d}{dx}[xy] + 2yy' &= -\sin(x) \\\\
\cos(xy)[y + xy'] + 2yy' &= -\sin(x) \\\\
\cos(xy)y + \cos(xy)xy' + 2yy' &= -\sin(x) \\\\
\cos(xy)xy' + 2yy' &= -\sin(x) - \cos(xy)y \\\\
[\cos(xy)x + 2y] \cdot y' &= -\sin(x) - \cos(xy)y \\\\
y' &= \frac{-\sin(x) - \cos(xy)y}{\cos(xy)x + 2y}
\end{align*}
x^3 + y^3 - 2xy = 0
. Find the slope of the tangent line at point
(1,1)
.
\begin{align*}
\frac{d}{dx}[x^3 + y^3 - 2xy] &= \frac{d}{dx}0 \\\\
3x^2 + 3y^2y' - 2[y + xy'] &= 0 \\\\
3x^2 + 3y^2y' - 2y - 2xy' &= 0 \\\\
3y^2y' - 2xy' &= -3x^2 + 2y \\\\
[3y^2 - 2x] \cdot y' &= -3x^2 + 2y \\\\
y' &= \frac{-3x^2 + 2y}{3y^2 - 2x} \\\\
\text{At }(1,1),\text{ } y' &= \frac{-3[1]^2 + 2[1]}{3[1]^2 - 2[1]} \\\\
\text{At }(1,1),\text{ } y' &= \frac{-3 + 2}{3 - 2} \\\\
\text{At }(1,1),\text{ } y' &= -1
\end{align*}
\ln[x+y] = x^2 + y^2
. Find
y'
.
\begin{align*}
\frac{d}{dx}\ln[x+y] &= \frac{d}{dx}[x^2 + y^2] \\\\
\frac{1}{x+y}\frac{d}{dx}[x+y] &= 2x + 2yy' \\\\
\frac{1}{x+y}[1+y'] &= 2x + 2yy' \\\\
\frac{1}{x+y} + \frac{y'}{x+y} &= 2x + 2yy' \\\\
\frac{y'}{x+y} - 2yy' &= 2x - \frac{1}{x+y} \\\\
y' \cdot [\frac{1}{x+y} - 2y] &= 2x - \frac{1}{x+y} \\\\
y' &= \frac{2x - \frac{1}{x+y}}{\frac{1}{x+y} - 2y} \\\\
y' &= \frac{\frac{2x^2 + 2xy}{x+y} - \frac{1}{x+y}}{\frac{1}{x+y} - \frac{2yx+2y^2}{x+y}} \\\\
y' &= \frac{\frac{2x^2 + 2xy-1}{x+y}}{\frac{1-2yx+2y^2}{x+y}} \\\\
y' &= \frac{2x^2 + 2xy-1}{1-2yx+2y^2}
\end{align*}