Back to the on-screen lesson ·
Differentiating a relation without solving it, and differentiating a function whose inverse has no formula.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to differentiate a relation implicitly — attaching a $\frac{dy}{dx}$ to every term containing $y$, using the product rule where both variables appear in one term, collecting and solving — evaluate the result at a point rather than at an $x$-value and say why that distinction is forced, read a vanishing denominator as a vertical tangent, and find the derivative of an inverse function at a point without any formula for the inverse.
$x^2 + y^2 = 25$ is not a function of $x$: the vertical line $x = 3$ meets it twice. Solving for $y$ gives two functions, $\pm\sqrt{25 - x^2}$, and choosing between them is an extra decision with no mathematical content.
Implicit differentiation avoids the choice. Differentiate both sides with respect to $x$, treating $y$ as an unnamed function of $x$. Every occurrence of $y$ is then an inner function, so the chain rule attaches a factor $\dfrac{dy}{dx}$:
$$2x + 2y\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{x}{y}.$$
There is no new rule here. Implicit differentiation is the chain rule applied with the discipline of remembering that $y$ depends on $x$.
The answer contains $y$. That is not an incompleteness to be fixed; it is the relation telling you that the slope depends on which branch you are on. An implicit derivative is evaluated at a point, never at an $x$-value.
The derivative of an inverse. Differentiate $f\bigl(f^{-1}(y)\bigr) = y$: $$f'\bigl(f^{-1}(y)\bigr)\cdot \bigl(f^{-1}\bigr)'(y) = 1 \quad\Longrightarrow\quad \bigl(f^{-1}\bigr)'(y) = \frac{1}{f'(x)} \ \text{ where } f(x) = y.$$ The evaluation point is the thing to get right: the reciprocal of $f'$ at the $x$ that maps to $y$, not at $y$.
That formula needs no formula for $f^{-1}$, which is what makes it useful: most invertible functions have no elementary inverse, and this differentiates them anyway.
Another way: picture
A circle with a tangent drawn at a point, and the radius drawn to the same point. The two are perpendicular, so their slopes multiply to $-1$: radius $y/x$, tangent $-x/y$. The implicit derivative reproduces a fact geometry already knew — and at the top and bottom of the circle, where the tangent is horizontal, and at the sides, where it is vertical and the formula's denominator vanishes.
Another way: steps
The step that goes wrong is a term containing both variables. $xy$ is a product, and its second factor is a function of $x$, so both rules apply: $$\frac{d}{dx}(xy) = 1 \cdot y + x\frac{dy}{dx} = y + x\frac{dy}{dx}.$$
For the folium $x^3 + y^3 = 6xy$:
$$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}.$$
Collect: $\bigl(3y^2 - 6x\bigr)\dfrac{dy}{dx} = 6y - 3x^2$, so $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.$$
Solving the original for $y$ would require the cubic formula and would produce three branches. Implicit differentiation handles all three at once and never mentions them, which is the whole point of it.
Forgetting the $\dfrac{dy}{dx}$. Differentiating $y^2$ to $2y$ rather than $2y\dfrac{dy}{dx}$ is the defining error, and it is the chain rule's missing factor wearing different clothes.
Treating $y$ as a constant. Then $\dfrac{d}{dx}(xy)$ comes out as $y$, and the product rule's second term is lost.
Evaluating at an $x$ alone. The slope formula usually contains $y$, and a relation can have two or more points with the same $x$.
Reading $(f^{-1})'(y) = 1/f'(y)$. The evaluation point is $x = f^{-1}(y)$. The two agree only when $f^{-1}(y) = y$, which is a coincidence.
Assuming a relation defines a function near every point. Where $\dfrac{dy}{dx}$ has a vanishing denominator it does not, and the tangent is vertical there. The formula reports this rather than hiding it.
An implicit derivative that still mentions $y$ looks incomplete, and the instinct is to substitute the relation back in and eliminate it. Usually that is impossible, and where it is possible it makes the answer worse.
The $y$ is carrying information. $-x/y$ says the slope at the top of the circle is $0$, at the sides undefined, and at $(3,4)$ and $(3,-4)$ opposite in sign — one formula describing every branch at once. Eliminating $y$ would mean choosing a branch, which is precisely the choice implicit differentiation was adopted to avoid.
The same is true of the inverse-function rule. $(f^{-1})'(y) = 1/f'(x)$ mentions $x$, and that is not a failure to simplify: it is the statement that you must know which input produced the output, which is what having an inverse means.
$x^2 + xy + y^2 = 7$ at $(1, 2)$. Differentiate term by term: $2x$, then the product rule on $xy$, then the chain rule on $y^2$.
Three terms, three different treatments.
$2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0$. Collect: $(x + 2y)\dfrac{dy}{dx} = -(2x + y)$.
Gather the unknown.
$\dfrac{dy}{dx} = -\dfrac{2x+y}{x+2y}$; at $(1,2)$ that is $-\dfrac{4}{5}$.
Both coordinates used.
Let $y = \arcsin x$, so $\sin y = x$ with $y \in [-\pi/2, \pi/2]$. Differentiate implicitly: $\cos y \dfrac{dy}{dx} = 1$.
Turn the inverse into a relation.
So $\dfrac{dy}{dx} = \dfrac{1}{\cos y}$, and $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ — positive, because $y$ is in the chosen range, and that is where the restriction from lesson 4 earns its keep.
The convention decides the sign.
$\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}$. An algebraic derivative for a trigonometric inverse, and no formula for $\arcsin$ was ever needed.
The left side is a product: $2xy + x^2\dfrac{dy}{dx} = 0$.
Product rule, then chain rule.
So $\dfrac{dy}{dx} = -\dfrac{2y}{x}$.
At $(2,2)$: $-2$. Check by solving — $y = 8x^{-2}$, so $y' = -16x^{-3}$, which at $x=2$ is $-2$. Agreed.
Match each statement to the derivative rule it expresses.
| $2y\,y'$ | $\cos(y)y'$ | $(f^{-1})'(b)=1/4$ | The inverse derivative formula does not give a finite slope | |
|---|---|---|---|---|
| Derivative of $y^2$ with respect to $x$ | ||||
| Derivative of $\sin y$ with respect to $x$ | ||||
| $f'(a)=4$ and $f(a)=b$ | ||||
| $f'(a)=0$ |
Match each differentiation fact to the conclusion it supplies.
| It becomes $2y\,dy/dx$ | Solve $2x+2y\,dy/dx=0$ for $dy/dx$ | Substitute both $x=3$ and $y=4$ | Use $1/f'(2)$, not $1/f'(8)$ | |
|---|---|---|---|---|
| Differentiate $y^2$ with respect to $x$ | ||||
| $x^2+y^2=25$ after differentiating | ||||
| Evaluate a relation's slope at $(3,4)$ | ||||
| Find $(f^{-1})'(8)$ when $f(2)=8$ |
The curve $x^2 + y^2 = 25$ passes through $(4, 3)$. Find $\dfrac{dy}{dx}$ there.
Answer:
Differentiating $y^{7}$ with respect to $x$ gives $7y^{6}\dfrac{dy}{dx}$. Which rule produces the extra factor?
$f(x) = 3x^3 + 8$ is increasing, so it has an inverse. Given $f(5) = 383$, find $(f^{-1})'(383)$.
Answer:
On the circle $x^2 + y^2 = 25$ the slope is $-x/y$. At how many points is the tangent vertical?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
On $x^2 + y^2 = 64 + 16$, the slope is $-x/y$. Why is it not enough to be told $x = 8$?
You can differentiate implicitly and find the derivative of an inverse. Say in your own words why the answer contains $y$ and why that is not a loose end. Next: the exponential and logarithmic derivatives, and the number $e$ they define.
8. Your turn: find $\dfrac{dy}{dx}$ for $x^2 y = 8$ at $(2, 2)$, step 3
Two routes where both are available.