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Implicit Differentiation

April 9th, 2009 |

Author: Merrin

An explicit relation between y and x can be written as

$y=f(x)$

An implicit relation between y and x can be written as

$g(y,x)=f(y,x)$

In an implicit relation no effort has been made to solve for y in terms of x or vice versa. Sometimes it is “impossible” to solve the relation. Interestingly we can still find the derivative of y from an implicit relation, just by taking the derivative of both sides of the equation.

$\cfrac{d}{dx} g(x,y)=\cfrac{d}{dx} f(x,y)$

By the chain rule, every time the derivative operator meets with a y a first derivative of y appears by the chain rule. Both sides therefore will be linear in the first derivative which can be solved. This new relation for the first derivative can be used to find the second derivative and so on

Example 1. Find the first derivative of y from the implicit relation

$x^{2} +9xy^{3} =x\sin y+\ln y$

Solution 1. Good luck solving the relation, but we can find y’

$\cfrac{d}{dx} (x^{2} +9xy^{3} )=\cfrac{d}{dx} (x\sin y+\ln y) \\2x+9y^{3} +27xy^{2} y'=\sin y+x(\cos y)y'+\cfrac{1}{y} y' \\y'\left(27xy^{2} -x\cos y-\cfrac{1}{y} \right)=\sin y-2x-9y^{3} \\y'=\cfrac{\sin y-2x-9y^{3} }{27xy^{2} -x\cos y-y^{-1} }$

Example 2. Relations with y and x are generally not functions. A simple relation is the circle. The circle does not pass the vertical line test so a circle is not a function, but the upper and lower branches are functions.

$x^{2} +y^{2} =1$

$y=\pm \sqrt{1-x^{2} }$

Depending on which sign chosen in front of the square root gives a particular branch, the lower semicircle or upper semicircle. These are explicit differentiable

relations.

We can find the equation of the tangent line at a particular point of the circle by using implicit differentiation at the point

$(2^{-1/2} ,2^{-1/2} )$

Solution 2. We follow the procedure and differentiate the implicit equation.

$\cfrac{d}{dx} (x^{2} +y^{2} )=\cfrac{d}{dx} 1 \\2x+2yy'=0 \\\cfrac{dy}{dx}=-\cfrac{x}{y}$

Now we plug in at the coordinate on the circle and we find

$\cfrac{d}{dx} y=-\cfrac{1/\sqrt{2} }{1/\sqrt{2} } =-1$

That is a symmetric location on the circle so we expect the slope to be negative one. The equation of the tangent line through the point on the circle is given by the point slope form of a line.

$y-y_{0} =f'(x_{0} )(x-x_{0} )$

${y=\cfrac{1}{\sqrt{2} } -\left(x-\cfrac{1}{\sqrt{2} } \right)} \\ {y=\sqrt{2} -x}$

We carefully choose a point which was on the circle. Choose another point say (5, 5) would give the same derivative. How is this possible? Well the derivative found by an implicit relation may correspond to the derivative for many different curves. In fact, by taking an arbitrary circle radius these relations all give the same implicit derivative. As a precaution, always check that where the derivative is evaluated is at a point on a branch of your relation, otherwise this definitely spells trouble.

Example 3. Find the implicit derivative of y given the following relation.

$a^{x} a^{y} =\ln (xy)$

Solution 3. We can simplify our work if we first take the logarithm of both sides.

${(x+y)\ln a}= {\ln (\ln (xy))} \\ {\cfrac{d}{dx} (x+y)\ln a}= {\cfrac{d}{dx} \ln (\ln (xy))} \\ {(1+y')\ln a} = {\cfrac{1}{\ln (xy)} \cfrac{1}{xy} (y+xy')}$

The final step is to algebraically solve for the derivative.

$\cfrac{dy}{dx} =\cfrac{\cfrac{1}{x\ln (xy)} -\ln a}{\ln a-\cfrac{1}{y\ln (xy)} } \cfrac{dy}{dx} =\cfrac{\cfrac{1}{x\ln (xy)} -\ln a}{\ln a-\cfrac{1}{y\ln (xy)} }$

Now I want to get a little bit artistic and show you how to different some nasty monsters or continued functions using implicit differentiation. An example of a continued function is

$f(x) = 1 + \cfrac{x} { 1+\cfrac{x} {1+...} }$

What you can do with continued functions is notice that there is a pattern that repeats and replace them with the symbol y.

$y = 1 + \cfrac{x} {y}$

Now we have an expression which can be differentiated with implicit differentiation. Before you do implicit differentiation it is sometimes helpful to do a little algebra so you don’t get bogged down with things like the quotient rule.

$y^2 = y + x$

$2yy' = y' + 1$

$y' = \cfrac{1}{2y-1}$

Example 4. Find the derivative of

$f(x) = \ln(x \ln (x ...$

Solution 4. We can write this expression as

$y = \ln(xy)$

Now we use implicit differentiation

$y' = \cfrac{y + xy'}{xy}$

Now the solution for y’ is just

$y' = \cfrac{y}{x(y-1)}$

When you start getting into continued function business you have to be aware where the function converges. Take the following series

$f(x) = 1 + x + x^2 + ...$

Using our trick we can rewrite this as

$y = 1 + x( 1+ x + x^2 ...) = y = 1+xy$

Or

$y = \cfrac{1}{1-x}$

Now obviously if x > 1 this equation is wrong, 1+2 + 4 + 8 … = infinity for example.

When working with your expressions for implicit derivatives you have to be extra careful. Another situation when implicit derivatives become handy are when you want to find the slopes of exotic curves.

Example 5. Analyze the slope along the curve of the folium of Descartes.

$x^3 + y^3 - 3axy = 0$

Solution 5. If you plot this function you find that is makes a loop dee loop. The curve crosses itself at the origin. Implicit differentiation is not going to work at that location, because the derivative is a function meaning it has a single value. All terms in implicit differentiation are proportional to the first derivative to the first power which can be solved for

$3x^2 +3y^2y' -3a(y+xy') = 0$