Implicit Differentiation

1. Motivating Example. Recall a circle consists of the points (x,y) satisfying
(1)x^{2}+y^{2}=1.
Question: how do we construct the line tangent to a point on the circle?

This problem really boils down to “How do we take the derivative of Equation (1)?” Since this is the meat of the matter, lets study this problem.

What we can try to do is differentiate everything in (1) with respect to x. So we get:
(2)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^{2}+y^{2})=\frac{\mathrm{d}1}{\mathrm{d}x}=0.
The right hand side vanishes (i.e., it’s zero), and the left hand side is a bit of trouble. We can use linearity
(3)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^{2}+y^{2})=\frac{\mathrm{d}}{\mathrm{d}x}(x^{2})+\frac{\mathrm{d}}{\mathrm{d}x}(y^{2}).
We could then use the chain rule
(4)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(x^{2})+\frac{\mathrm{d}}{\mathrm{d}x}(y^{2}) = 2x + \frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}y}(y^{2}).
Then evaluate the derivative to obtain
(5)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^{2}+y^{2})=2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0.
Simple algebraic manipulations give us
(6)\displaystyle \frac{-x}{y}=\frac{\mathrm{d}y}{\mathrm{d}x}.
So differentiating everything with respect to x produces the desired result, after some elementary manipulation.

This technique (differentiate everything, then do some cleaning up) is precisely what people call “Implicit Differentiation”. It relies on the chain rule in a critical way!

Example. Consider the expression
\displaystyle (x+y)^{n}=x^{n}+nx^{n-1}y+\dots+y^{n}
What is its derivative with respect to x? We see that
\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(x+y)^{n}=n(x+y)^{n-1}\frac{\mathrm{d}(x+y)}{\mathrm{d}x}
Great…what now? Well, how about use linearity to write
\displaystyle \frac{\mathrm{d}(x+y)}{\mathrm{d}x}=\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{\mathrm{d}y}{\mathrm{d}x}=1+\frac{\mathrm{d}y}{\mathrm{d}x}
This works by the chain rule. Then we plug this back in, obtaining
\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(x+y)^{n}=n(x+y)^{n-1}\left(1+\frac{\mathrm{d}y}{\mathrm{d}x}\right)
for the derivative.

Example (Power Rule Plus Plus). We had discussed the power rule for integer exponents. That is, x^{n} for n = 0, \pm1, \pm2, etc. What if we try differentiating x^{q} where q=m/n is a rational number?

Lets write
(7)\displaystyle y^{n}=x^{m}\quad\mbox{or}\quad y=x^{m/n}
Now lets take its derivative. Before we didn’t know what to do, but NOW we have implicit differentiation! So we can differentiate everything with respect to x getting:
(8)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}y^{n}=\frac{\mathrm{d}}{\mathrm{d}x}x^{m}=mx^{m-1}
Implicit differentiation enables us to compute the left hand side as
(9)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}y^{n}=\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}y}y^{n}=ny^{n-1}\frac{\mathrm{d}y}{\mathrm{d}x}
We plug this back in:
(10)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}y^{n}=ny^{n-1}\frac{\mathrm{d}y}{\mathrm{d}x}  =mx^{m-1}
Divide through by ny^{n-1} and we obtain
(11)\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}  =\frac{mx^{m-1}}{ny^{n-1}}=\frac{mx^{m-1}}{n(x^{m/n})^{n-1}}
Now we just do some algebra to deduce
(12)\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}  =\frac{m}{n}x^{m-1-(n-1)(m/n)}
Lets observe the exponent simplifies to
(13)\displaystyle m-1-(n-1)(m/n) = m-1-n(m/n)+(m/n) = (m/n)-1
Thus we can find
(14)\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}  =\frac{m}{n}x^{(m/n)-1}
This could only be done through the chain rule and implicit differentiation!

Reiterating this example: if q is a rational number, then
(15)\displaystyle \frac{\mathrm{d}x^{q}}{\mathrm{d}x}  =qx^{q-1}
This is the power rule extended to rational powers.

Lets give the reader some exercises to solidify their command of this technique.

Exercise 1 (Kampyle of Eudoxus). Consider the curve x^4=x^2+y^2, called the Kampyle of Eudoxus. Suppose (x_{0},y_{0}) lies on the curve. What is the tangent line to the Kampyle of Eudoxus at (x_{0},y_{0})?

Exercise 2. Consider the Lemniscate of Bernoulli, which in Cartesian coordinates is (x^2 + y^2)^2 = 2a^2 (x^2 - y^2)\,; find the tangent line passing through the point (x_0,y_0).

Exercise 3. The Tschinhausen cubic is a curve y^{2}=x^{3}+3x^{2}. Find the tangent line passing through the point (x_0,y_0).

Exercise 4. Consider \displaystyle xy=\cos(x-y). What is \mathrm{d}y/\mathrm{d}x?

Exercise 5. Recall a circle is defined by x^{2}+y^{2}=1. Calculate
\displaystyle f(x,y)=\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}
When will f(x,y)>0 and when will f(x,y)<0. [Hint: f(x,y) should be rewritten so it depends only on y.]

Exercise 6. Given the equation
\displaystyle y = x + x\mathrm{e}^{y}
find \mathrm{d}y/\mathrm{d}x. [From Math.SE]

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
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4 Responses to Implicit Differentiation

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