[Figure: FiveStudy logo]
FiveStudy
Calculus BC · Unit 3 · 3.2
AP Calculus · Differentiation: Composite, Implicit, and Inverse Functions
Implicit Differentiation
Never solve for $y$. Differentiate both sides and let every $y$ carry its chain-rule factor.
key idea$\text{Differentiate BOTH sides; every }y\text{ contributes a }\tfrac{dy}{dx}$
The procedure
Some relations cannot be solved for $y$ — $x^{2}+y^{2}=25$ has two branches, and $\sin(xy)=x$ has no closed form at all. Implicit differentiation finds $\tfrac{dy}{dx}$ without ever isolating $y$, by treating $y$ as an unnamed function of $x$.
1
Differentiate both sides with respect to $x$
Apply $\dfrac{d}{dx}$ to every term on both sides of the equation. The equals sign stays.
in a relation out an equation containing $\tfrac{dy}{dx}$
2
Every $y$ picks up a chain-rule factor
$y$ is a function of $x$, so $\dfrac{d}{dx}\left[y^{3}\right]=3y^{2}\dfrac{dy}{dx}$. A term in $x$ alone differentiates normally. This step is the whole technique.
watch every $y$ term
3
Use the product rule on mixed terms
A term like $xy$ contains both variables, so it needs the product rule: $\dfrac{d}{dx}[xy]=1\cdot y+x\dfrac{dy}{dx}$.
watch terms with both $x$ and $y$
4
Collect every $\tfrac{dy}{dx}$ on one side
Move all terms containing $\dfrac{dy}{dx}$ to the left and everything else to the right. This is ordinary algebra — treat $\dfrac{dy}{dx}$ as a single symbol.
in a mixed equation out a factored one
5
Factor and divide
Factor out $\dfrac{dy}{dx}$, then divide. The answer usually contains both $x$ and $y$ — that is expected, not a mistake.
out $\dfrac{dy}{dx}$ in terms of $x$ and $y$
| Term | Derivative with respect to $x$ | Why |
|---|
| $x^{3}$ | $3x^{2}$ | ordinary power rule |
| $y^{3}$ | $3y^{2}\dfrac{dy}{dx}$ | chain rule — $y$ depends on $x$ |
| $xy$ | $y+x\dfrac{dy}{dx}$ | product rule |
| $\sin y$ | $\cos y\dfrac{dy}{dx}$ | chain rule |
| $x^{2}y^{3}$ | $2xy^{3}+3x^{2}y^{2}\dfrac{dy}{dx}$ | product and chain |
| $7$ | $0$ | constant |
Row 2 against row 1 is the entire difference. If a $\tfrac{dy}{dx}$ is missing from a $y$ term, the rest of the work cannot recover.
Worked examples
EX 1
The circle
$x^{2}+y^{2}=25$Find $\dfrac{dy}{dx}$ for $x^{2}+y^{2}=25$.
1
Differentiate both sides
$\dfrac{d}{dx}\left[x^{2}\right]+\dfrac{d}{dx}\left[y^{2}\right]=\dfrac{d}{dx}[25]$.
2
Apply the chain rule to the $y$ term
$2x+2y\dfrac{dy}{dx}=0$.
3
Isolate
$2y\dfrac{dy}{dx}=-2x$.
4
Divide
$\dfrac{dy}{dx}=-\dfrac{x}{y}$.
5
Read the geometry
At $(3,4)$ the slope is $-\tfrac34$; at $(3,-4)$ it is $+\tfrac34$. One formula covers both branches — which is why the answer must contain $y$.
EX 2
A mixed term
$x^{2}+xy+y^{2}=7$Find $\dfrac{dy}{dx}$ for $x^{2}+xy+y^{2}=7$.
1
Differentiate term by term
$2x+\dfrac{d}{dx}[xy]+2y\dfrac{dy}{dx}=0$.
2
Product rule on $xy$
$\dfrac{d}{dx}[xy]=y+x\dfrac{dy}{dx}$.
3
Substitute
$2x+y+x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0$.
4
Collect and factor
$\dfrac{dy}{dx}(x+2y)=-(2x+y)$.
5
Divide
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$.
EX 3
Tangent line at a point
$x^{3}+y^{3}=9xy$Find the tangent to $x^{3}+y^{3}=9xy$ at $(2,4)$.
1
Differentiate
$3x^{2}+3y^{2}\dfrac{dy}{dx}=9\left(y+x\dfrac{dy}{dx}\right)$.
2
Expand and collect
$3x^{2}+3y^{2}\dfrac{dy}{dx}=9y+9x\dfrac{dy}{dx}$, so $\dfrac{dy}{dx}\left(3y^{2}-9x\right)=9y-3x^{2}$.
3
Solve
$\dfrac{dy}{dx}=\dfrac{9y-3x^{2}}{3y^{2}-9x}=\dfrac{3y-x^{2}}{y^{2}-3x}$.
4
Substitute the point
At $(2,4)$: $\dfrac{12-4}{16-6}=\dfrac{8}{10}=\dfrac45$.
5
Write the line
$y-4=\dfrac45(x-2)$. Substituting the point after solving is far less work than carrying numbers through.
EX 4
Vertical tangents
where the denominator vanishesFor the circle $x^{2}+y^{2}=25$, where is the tangent vertical?
1
Recall the derivative
$\dfrac{dy}{dx}=-\dfrac{x}{y}$.
2
Ask when a slope is undefined
A vertical tangent has no slope, which happens when the denominator is zero: $y=0$.
3
Find those points
With $y=0$ the relation gives $x^{2}=25$, so $x=\pm5$.
4
State them
Vertical tangents at $(5,0)$ and $(-5,0)$ — the left and right extremes of the circle.
5
The general habit
Setting the numerator to zero finds horizontal tangents; setting the denominator to zero finds vertical ones.
Where it goes wrong
A $y$ term without $\tfrac{dy}{dx}$
$\dfrac{d}{dx}\left[y^{2}\right]$ is $2y\dfrac{dy}{dx}$, not $2y$. Every single $y$ carries the factor — this is the error that ends the problem.
Terms with both variables need the product rule
$xy$, $x^{2}y$ and $x\sin y$ are products. Differentiating them as though only one variable moved drops half the derivative.
The answer containing $y$ is normal
$\dfrac{dy}{dx}=-\dfrac{x}{y}$ is a complete answer. The relation defines several branches, and $y$ is what tells them apart.
Substitute the point last
Solve for $\tfrac{dy}{dx}$ symbolically, then put the numbers in. Substituting early turns a clean algebra step into a mess.