Tag Archives: implicit derivative

Implicit Derivative Question

Yesterday, this question posted within the AP Calculus Community about implicit derivatives.

Below, I argue why the derivatives MUST be the same, show how four different variations can all be shown to give the same derivative, and provide a final conclusion.

INITIAL INTUITION

The Desmos graph of the given relation, is $y^2 = \frac{x-1}{x+1}$, is shown below.  Logically, it seems that even when the terms of the relation are algebraically rearranged, the graph should be invariant.   The other two forms mentioned in the Community post are on lines 2 and 3.  Lines 4, 5, and 6 show three other variations.  Here is the link to my Desmos graph allowing you to change between the forms to visually confirm the graphical invariance intuition.

If calculus “works”, it also shouldn’t matter how one calculates a derivative.   While the forms of the derivative certainly could LOOK different,  because any point on the invariant graph has the same tangent line no matter what the form of its equation, and the derivative of a relation at a point is the slope of that invariant tangent line, then the derivative also MUST be invariant.

CALCULATING “DIFFERENT” DERIVATIVES

To show the derivatives are fundamentally all the same (as suspected by the initial post), I calculate the derivatives of the equations on lines 1 and 3 given in the initial post as well as my variations on lines 4 and 6.

LINE 1:

Using the Chain Rule on the left and the Quotient Rule on the right gives

LINE 3:

This version is more complicated, requiring the Product Rule in addition to the earlier Chain and Quotient Rules.  In the penultimate line, I used the original equation to substitute for $y^2$ to transform the derivative into the same form as line 1.

LINE 4:

This time, differentiation requires only the Chain and Product Rules.

After the usual substitution for $y^2$, I multiplied both sides by $(x+1)$ to clear the denominator and solved for $y'$, returning the same result.

LINE 6:

This time, the relation is solved for x, resulting in a much more complicated Quotient+Chain Rule calculation, but substituting for $y^2$ and changing the form leads once again to the same answer.

Hopefully this is convincing evidence that all derivative forms can be shown to be equivalent.   If you’re still learning implicit differentiation, I encourage you to show the derivatives from the lines 2 and 5 variations are also equivalent.

CONCLUSION

So which approach is “best”?  In my opinion, it all depends on your personal comfort with algebraic manipulations.  Some prefer to just take a derivative from the given form of $y^2 = \frac{x-1}{x+1}$.  I avoid the more complicated quotient rule whenever I can, so the variation from line 4 would have been my approach.

The cool part is that it doesn’t matter what approach you use, so long as your algebraic manipulations are sound.  You don’t have to accept the form in which a problem is given; CHANGE IT to a form that works for you!

Many Roads Give Same Derivative

A recent post in the AP Calculus Community expressed some confusion about different ways to compute $\displaystyle \frac{dy}{dx}$ at (0,4) for the function $x=2ln(y-3)$.  I share below the two approaches suggested in the original post, proffer two more, and a slightly more in-depth activity I’ve used in my calculus classes for years.  I conclude with an alternative to derivatives of inverses.

Two Approaches Initially Proposed

1 – Accept the function as posed and differentiate implicitly.

$\displaystyle \frac{d}{dx} \left( x = 2 ln(y-3) \right)$

$\displaystyle 1 = 2*\frac{1}{y-3} * \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{y-3}{2}$

Which gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ at (x,y)=(0,4).

2 – Solve for y and differentiate explicitly.

$\displaystyle x = 2ln(y-3) \longrightarrow y = 3 + e^{x/2}$

$\displaystyle \frac{dy}{dx} = e^{x/2} * \frac{1}{2}$

Evaluating this at (x,y)=(0,4) gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ .

Two Alternative Approaches

3 – Substitute early.

The question never asked for an algebraic expression of $\frac{dy}{dx}$, only the numerical value of this slope.  Because students tend to make more silly mistakes manipulating algebraic expressions than numeric ones, the additional algebra steps are unnecessary, and potentially error-prone.  Admittedly, the manipulations are pretty straightforward here, in more algebraically complicated cases, early substitutions could significantly simplify work. Using approach #1 and substituting directly into the second line gives

$\displaystyle 1 = 2 * \frac{1}{y-3} * \frac{dy}{dx}$.

At (x,y)=(0,4), this is

$\displaystyle 1 = 2 * \frac{1}{4-3}*\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}$

The numeric manipulations on the right side are obviously easier than the earlier algebra.

4 – Solve for $\frac{dx}{dy}$ and reciprocate.

There’s nothing sacred about solving for $\frac{dy}{dx}$ directly.  Why not compute the derivative of the inverse and reciprocate at the end? Differentiating first with respect to y eventually leads to the same solution.

$\displaystyle \frac{d}{dy} \left( x = 2 ln(y-3) \right)$

$\displaystyle \frac{dx}{dy} = 2 * \frac{1}{y-3}$

At (x,y)=(0,4), this is

$\displaystyle \frac{dx}{dy} = \frac{2}{4-3} = 2$, so

$\displaystyle \frac{dy}{dx} = \frac{1}{2}$.

Equivalence = A fundamental mathematical concept

I sometimes wonder if teachers should place much more emphasis on equivalence.  We spend so much time manipulating expressions in mathematics classes at all levels, changing mathematical objects (shapes, expressions, equations, etc.) into a different, but equivalent objects.  Many times, these manipulations are completed under the guise of “simplification.”  (Here is a brilliant Dan Teague post cautioning against taking this idea too far.)

But it is critical for students to recognize that proper application of manipulations creates equivalent expressions, even if when the resulting expressions don’t look the same.   The reason we manipulate mathematical objects is to discover features about the object in one form that may not be immediately obvious in another.

For the function $x = 2 ln(y-3)$, the slope at (0,4) must be the same, no matter how that slope is calculated.  If you get a different looking answer while using correct manipulations, the final answers must be equivalent.

Another Example

A similar question appeared on the AP Calculus email list-server almost a decade ago right at the moment I was introducing implicit differentiation.  A teacher had tried to find $\displaystyle \frac{dy}{dx}$ for

$\displaystyle x^2 = \frac{x+y}{x-y}$

using implicit differentiation on the quotient, manipulating to a product before using implicit differentiation, and finally solving for y in terms of x to use an explicit derivative.

1 – Implicit on a quotient

Take the derivative as given:\$

$\displaystyle \frac{d}{dx} \left( x^2 = \frac{x+y}{x-y} \right)$

$\displaystyle 2x = \frac{(x-y) \left( 1 + \frac{dy}{dx} \right) - (x+y) \left( 1 - \frac{dy}{dx} \right) }{(x-y)^2}$

$\displaystyle 2x * (x-y)^2 = (x-y) + (x-y)*\frac{dy}{dx} - (x+y) + (x+y)*\frac{dy}{dx}$

$\displaystyle 2x * (x-y)^2 = -2y + 2x * \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{-2x * (x-y)^2 + 2y}{2x}$

2 – Implicit on a product

Multiplying the original equation by its denominator gives

$x^2 * (x - y) = x + y$ .

Differentiating with respect to x gives

$\displaystyle 2x * (x - y) + x^2 * \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$

$\displaystyle 2x * (x-y) + x^2 - 1 = x^2 * \frac{dy}{dx} + \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{2x * (x-y) + x^2 - 1}{x^2 + 1}$

3 – Explicit

Solving the equation at the start of method 2 for y gives

$\displaystyle y = \frac{x^3 - x}{x^2 + 1}$.

Differentiating with respect to x gives

$\displaystyle \frac{dy}{dx} = \frac {\left( x^2+1 \right) \left( 3x^2 - 1\right) - \left( x^3 - x \right) (2x+0)}{\left( x^2 + 1 \right) ^2}$

Equivalence

Those 3 forms of the derivative look VERY DIFFERENT.  Assuming no errors in the algebra, they MUST be equivalent because they are nothing more than the same derivative of different forms of the same function, and a function’s rate of change doesn’t vary just because you alter the look of its algebraic representation.

Substituting the y-as-a-function-of-x equation from method 3 into the first two derivative forms converts all three into functions of x.  Lots of by-hand algebra or a quick check on a CAS establishes the suspected equivalence.  Here’s my TI-Nspire CAS check.

Here’s the form of this investigation I gave my students.

Final Example

I’m not a big fan of memorizing anything without a VERY GOOD reason.  My teachers telling me to do so never held much weight for me.  I memorized as little as possible and used that information as long as I could until a scenario arose to convince me to memorize more.  One thing I managed to avoid almost completely were the annoying derivative formulas for inverse trig functions.

For example, find the derivative of $y = arcsin(x)$ at $x = \frac{1}{2}$.

Since arc-trig functions annoy me, I always rewrite them.  Taking sine of both sides and then differentiating with respect to x gives.

$sin(y) = x$

$\displaystyle cos(y) * \frac{dy}{dx} = 1$

I could rewrite this equation to give $\frac{dy}{dx} = \frac{1}{cos(y)}$, a perfectly reasonable form of the derivative, albeit as a less-common  expression in terms of y.  But I don’t even do that unnecessary algebra.  From the original function, $x=\frac{1}{2} \longrightarrow y=\frac{\pi}{6}$, and I substitute that immediately after the differentiation step to give a much cleaner numeric route to my answer.

$\displaystyle cos \left( \frac{\pi}{6} \right) * \frac{dy}{dx} = 1$

$\displaystyle \frac{\sqrt{3}}{2} * \frac{dy}{dx} = 1$

$\displaystyle \frac{dy}{dx} = \frac{2}{\sqrt{3}}$

And this is the same result as plugging $x = \frac{1}{2}$ into the memorized version form of the derivative of arcsine.  If you like memorizing, go ahead, but my mind remains more nimble and less cluttered.

One final equivalent approach would have been differentiating $sin(y) = x$ with respect to y and reciprocating at the end.

CONCLUSION

There are MANY ways to compute derivatives.  For any problem or scenario, use the one that makes sense or is computationally easiest for YOU.  If your resulting algebra is correct, you know you have a correct answer, even if it looks different.  Be strong!