Day 8: Implicit & Exponential/Log Derivatives

Implicit Differentiation

Not every curve can be solved explicitly for $y$ in terms of $x$ — circles, folium-shaped curves, and many AP FRQ setups only give you an equation relating $x$ and $y$ . Implicit differentiation lets you find $\frac{dy}{dx}$ anyway, by differentiating both sides of the equation as-is.

Core Theorem

Differentiate each term of the equation with respect to

x

. Every time you differentiate a term containing

y

, treat

y

as a function of

x

and multiply by

\frac{dy}{dx}

(the chain rule). For example,

\frac{d}{dx}(xy) = y + x\frac{dy}{dx}

(product rule) and

\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}

(chain rule).

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find

\frac{dy}{dx}

for the circle

x^2+y^2=25

, and use it to find the tangent line at the point

(3,-4)

.

How to think about it

When you cannot solve for

y

, differentiate both sides as they are. Every time you differentiate a

y

term, the chain rule tacks on a

\frac{dy}{dx}

; then collect those terms and solve for it.

Key idea: Differentiate both sides with respect to $x$ ; the $y^2$ term needs the chain rule.

Full solution

1. Step 1: Differentiate Both Sides

2x + 2y\frac{dy}{dx} = 0

.

2. Step 2: Solve for dy/dx

\frac{dy}{dx} = -\frac{x}{y}

.

3. Step 3: Evaluate the Slope at (3, -4)

\frac{dy}{dx}\Big|_{(3,-4)} = -\frac{3}{-4} = \frac{3}{4}

.

4. Step 4: Write the Tangent Line

y - (-4) = \frac{3}{4}(x-3) \implies y = \frac{3}{4}x - \frac{25}{4}

.

Worked Example 02Medium

Find

y'

for the folium of Descartes:

x^3+y^3=6xy

.

How to think about it

Key idea: The right side $6xy$ needs the product rule, and every $y$ term needs an extra $y'$ tacked on.

Full solution

1. Step 1: Differentiate Both Sides

3x^2 + 3y^2y' = 6y + 6xy'

(the right side used the product rule on

6xy

).

2. Step 2: Collect All y' Terms on One Side

3y^2y' - 6xy' = 6y - 3x^2 \implies y'(3y^2-6x) = 6y-3x^2

.

3. Step 3: Solve for y'

y' = \frac{6y-3x^2}{3y^2-6x} = \frac{2y-x^2}{y^2-2x}

.

Step-by-Step SOP

1
Differentiate Every Term w.r.t. x
Go term by term. For $x$ -only terms, differentiate normally. For $y$ -only or mixed terms, apply the chain rule (and product rule, if mixed) and attach $y'$ .
2
Collect and Factor
Move every term containing $y'$ to one side of the equation, factor $y'$ out, and solve for it.
3
Substitute the Point Last
Keep $y'$ in terms of both $x$ and $y$ until the very end, then plug in the specific point you care about.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

For the curve

x^2+xy+y^3=1

, find

y'

and

y''

at the point

(1,0)

.

Need a hint?

First confirm

(1,0)

is actually on the curve, then differentiate once for

y'

, and differentiate that whole equation again for

y''

.

Show solution

1. Step 1: Differentiate Once for y'

2x + (y+xy') + 3y^2y' = 0 \implies y'(x+3y^2) = -(2x+y) \implies y' = -\frac{2x+y}{x+3y^2}

.

2. Step 2: Evaluate y' at (1, 0)

y'(1,0) = -\frac{2(1)+0}{1+3(0)^2} = -\frac{2}{1} = -2

.

3. Step 3: Differentiate the y' Equation Again for y''

Differentiating

2x+y+xy'+3y^2y'=0

term by term:

2 + y' + (y'+xy'') + (6y(y')^2+3y^2y'') = 0

, which simplifies to

y''(x+3y^2) = -\left[2+2y'+6y(y')^2\right]

.

4. Step 4: Substitute x=1, y=0, y'=-2

y''(1+0) = -[2+2(-2)+6(0)(4)] = -[2-4+0] = 2

, so

y''(1,0) = 2

.

Practice 02AB/BC Standard

Verify that

y = ax^3

and

x^2+3y^2=b

form a family of orthogonal trajectories (every curve in one family meets every curve in the other at a right angle).

Need a hint?

Find the slope of each family in terms of

x

and

y

only (eliminate the parameters

a

and

b

), then multiply the two slopes together.

Show solution

1. Step 1: Find the Slope of y = ax³ in Terms of x, y

y' = 3ax^2

. Since

a = \frac{y}{x^3}

from the original equation,

y' = 3\left(\frac{y}{x^3}\right)x^2 = \frac{3y}{x}

.

2. Step 2: Find the Slope of x² + 3y² = b Implicitly

2x+6yy'=0 \implies y' = -\frac{x}{3y}

.

3. Step 3: Multiply the Two Slopes

\frac{3y}{x} \cdot \left(-\frac{x}{3y}\right) = -1

.

4. Step 4: Conclude

Since the product of the slopes is

-1

at every intersection point, the two families are orthogonal trajectories.

Practice 03Hard

Find all points on the curve

x^2 - xy + y^2 = 3

where the tangent line is horizontal.

Need a hint?

A horizontal tangent means

y'=0

. Differentiate implicitly, set the numerator of

y'

equal to 0, then substitute back into the original equation to find the actual points.

Show solution

1. Step 1: Differentiate Implicitly

2x - (y+xy') + 2yy' = 0 \implies y'(2y-x) = y-2x \implies y' = \frac{y-2x}{2y-x}

.

2. Step 2: Set the Numerator to Zero

y - 2x = 0 \implies y = 2x

.

3. Step 3: Substitute Back into the Original Equation

x^2 - x(2x) + (2x)^2 = 3 \implies x^2-2x^2+4x^2=3 \implies 3x^2=3 \implies x=\pm 1

.

4. Step 4: State the Points

At

x=1

:

y=2

. At

x=-1

:

y=-2

. Both denominators (

2y-x

) are nonzero at these points, confirming horizontal tangents at

(1,2)

and

(-1,-2)

.

Common Pitfalls

⚠
Every y Needs Its Own dy/dxIt's easy to differentiate $y^3$ correctly as $3y^2y'$ but then forget to attach $y'$ to a simpler term like a lone $y$ later in the same equation — track every single $y$ term.
⚠
Mixed Terms Like xy Need the Product Rule $\frac{d}{dx}(xy) = y + xy'$ , NOT just $y'$ or just $x y'$ — since $x$ and $y$ are both functions of $x$ here, the product rule is mandatory.
⚠
Solve for dy/dx Only at the EndDon't try to isolate $y'$ mid-differentiation. Differentiate the entire equation first, then collect all $y'$ terms on one side and factor.

Derivatives of Exponential & Logarithmic Functions

$e^x$ is the only function that is its own derivative — every other exponential or logarithmic derivative picks up an extra factor of $\ln(\text{base})$ . When a variable shows up in BOTH the base and the exponent, none of the usual rules apply directly — that's when logarithmic differentiation comes in.

Core Theorem

\frac{d}{dx}e^x = e^x

\frac{d}{dx}a^x = a^x \ln a

\frac{d}{dx}\ln x = \frac{1}{x}\ (x>0)

, and

\frac{d}{dx}\ln|x| = \frac{1}{x}\ (x \neq 0)

\frac{d}{dx}\log_a x = \frac{1}{x\ln a}

General power rule:

\frac{d}{dx}u^r = ru^{r-1}\frac{du}{dx}

for any real

r

.
Logarithmic differentiation: to differentiate

y=[f(x)]^{g(x)}

, take

\ln

of both sides first:

\ln y = g(x)\ln f(x)

, then differentiate implicitly.

Watch the TikTok ExplanationDerivative of Exponential Functions→Watch the TikTok ExplanationDerivative of Logarithmic Functions→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Differentiate

y = e^{\sec(3\theta)}

.

How to think about it

\frac{d}{dx}e^{u} = e^{u}\cdot u'

— the

e^{u}

comes back unchanged and you multiply by the derivative of the exponent. If that exponent is itself composite, it needs its own chain rule.

Key idea: The outside function is $e^{(\cdot)}$ — apply the chain rule and don't forget the derivative of $\sec(3\theta)$ needs its own chain rule too.

Full solution

1. Step 1: Differentiate the Outer e^(...) Layer

\frac{dy}{d\theta} = e^{\sec(3\theta)} \cdot \frac{d}{d\theta}\left[\sec(3\theta)\right]

.

2. Step 2: Differentiate sec(3θ) with the Chain Rule

\frac{d}{d\theta}\sec(3\theta) = \sec(3\theta)\tan(3\theta) \cdot 3

.

3. Step 3: Combine

\frac{dy}{d\theta} = 3\sec(3\theta)\tan(3\theta)\,e^{\sec(3\theta)}

.

Worked Example 02Medium

Differentiate

f(x) = \sqrt{\ln x}

.

How to think about it

Key idea: Rewrite as $(\ln x)^{1/2}$ and apply the chain rule, recalling that the derivative of $\ln x$ is $\frac{1}{x}$ .

Full solution

1. Step 1: Rewrite as a Power

f(x) = (\ln x)^{1/2}

.

2. Step 2: Differentiate the Outer Power

f'(x) = \frac{1}{2}(\ln x)^{-1/2} \cdot \frac{d}{dx}(\ln x)

.

3. Step 3: Multiply by the Inner Derivative

f'(x) = \frac{1}{2}(\ln x)^{-1/2} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}

.

Step-by-Step SOP

1
Classify the Expression
Determine whether the variable is in the base only (power rule), the exponent only (exponential rule), or both (logarithmic differentiation needed).
2
For Logarithmic Differentiation, Take ln of Both Sides First
Rewrite $\ln y = g(x)\ln f(x)$ , differentiate implicitly, then multiply both sides by the original $y$ to isolate $y'$ .
3
Don't Forget the Chain Rule for Nested Arguments
Even after picking the right base rule, if the exponent or argument is itself a function of $x$ , you still need to multiply by its derivative.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01AB/BC Standard

Use logarithmic differentiation to find

\frac{dy}{dx}

for

y = x^{\sqrt{x}}

.

Need a hint?

Since both the base AND the exponent contain

x

, no ordinary rule applies directly — take

\ln

of both sides first.

Show solution

1. Step 1: Take the Natural Log of Both Sides

\ln y = \sqrt{x}\ln x

.

2. Step 2: Differentiate Both Sides Implicitly

\frac{y'}{y} = \frac{1}{2\sqrt{x}}\ln x + \sqrt{x}\cdot\frac{1}{x} = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}

.

3. Step 3: Combine the Right Side

\frac{y'}{y} = \frac{\ln x + 2}{2\sqrt{x}}

.

4. Step 4: Solve for y' by Multiplying by y

y' = x^{\sqrt{x}}\cdot\frac{\ln x + 2}{2\sqrt{x}}

.

Practice 02Medium

Compare the derivatives of four similar-looking expressions: (a)

5^3

, (b)

x^3

, (c)

2^{\sin x}

, (d)

x^x

.

Need a hint?

Ask which part is the variable: the base, the exponent, both, or neither?

Show solution

1. (a) Constant Base, Constant Exponent

5^3

is just the number 125 — a constant. Its derivative is

0

.

2. (b) Variable Base, Constant Exponent → Power Rule

\frac{d}{dx}(x^3) = 3x^2

.

3. (c) Constant Base, Variable Exponent → Exponential Rule

\frac{d}{dx}\left(2^{\sin x}\right) = 2^{\sin x}(\ln 2)\cos x

.

4. (d) Variable Base AND Variable Exponent → Logarithmic Differentiation

\frac{d}{dx}(x^x) = x^x(\ln x + 1)

, found by taking

\ln y = x\ln x

and differentiating implicitly.

Practice 03Easy

At which point on the curve

y=e^x

is the tangent line parallel to

y=2x

?

Need a hint?

'Parallel' means the slopes match. Set

y'=e^x

equal to the slope of the given line and solve for

x

.

Show solution

1. Step 1: Set the Derivative Equal to the Target Slope

y' = e^x = 2

.

2. Step 2: Solve for x

x = \ln 2

.

3. Step 3: Find the y-Coordinate

y = e^{\ln 2} = 2

.

4. Step 4: State the Point

The tangent line is parallel to

y=2x

at the point

(\ln 2,\ 2)

.

Common Pitfalls

⚠
Only e^x Reproduces Itself $\frac{d}{dx}a^x = a^x\ln a$ , NOT just $a^x$ . The extra $\ln a$ factor is only invisible when $a=e$ , since $\ln e = 1$ .
⚠
Identify Which Part Is the Variable Before Choosing a RulePower rule, exponential rule, and logarithmic differentiation look similar but apply to different situations — always check whether the variable is in the base, the exponent, or both before picking a method.

Higher-Order Derivatives

Differentiating a derivative gives the second derivative, differentiating again gives the third, and so on. The second derivative is the workhorse — it powers concavity, the Second Derivative Test, and acceleration in particle motion — so being fluent at $f''$ (including implicitly) matters.

Core Theorem

Notation:

f'(x)

,

f''(x)

,

f'''(x)

,

f^{(4)}(x), \dots

or

\dfrac{dy}{dx}, \dfrac{d^2y}{dx^2}, \dfrac{d^3y}{dx^3}, \dots

Each one is just the derivative of the one before it. For an implicit curve, differentiate

\dfrac{dy}{dx}

again with respect to

x

(using the quotient/product and chain rules), then substitute the expression for

\dfrac{dy}{dx}

back in.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

For

f(x) = x^4 - 3x^3 + 2x - 7

, find

f''(x)

and

f'''(x)

.

How to think about it

f''

is just the derivative of

f'

— keep going. Simplify each derivative before taking the next one, or the algebra snowballs.

Key idea: Differentiate once for $f'$ , again for $f''$ , again for $f'''$ .

Full solution

1. Step 1: First Derivative

f'(x) = 4x^3 - 9x^2 + 2

.

2. Step 2: Second Derivative

f''(x) = 12x^2 - 18x

.

3. Step 3: Third Derivative

f'''(x) = 24x - 18

.

Worked Example 02Medium

If

y = \sin(2x)

, find

\dfrac{d^2y}{dx^2}

.

How to think about it

Key idea: Each derivative brings down a factor of 2 from the chain rule.

Full solution

1. Step 1: First Derivative

\dfrac{dy}{dx} = 2\cos(2x)

.

2. Step 2: Second Derivative

\dfrac{d^2y}{dx^2} = -4\sin(2x)

.

Step-by-Step SOP

1
Differentiate, Then Differentiate the Result
$f''$ is literally the derivative of $f'$ . Simplify $f'$ before differentiating again to avoid a mess.
2
For Implicit Second Derivatives, Substitute Last
Find $\dfrac{dy}{dx}$ , differentiate it again with respect to $x$ , then replace every remaining $\dfrac{dy}{dx}$ with the expression you found in step 1.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

A particle's position is

s(t) = t^3 - 6t^2 + 9t

. Find its acceleration at

t = 2

.

Need a hint?

Velocity is

s'(t)

; acceleration is

s''(t)

.

Show solution

1. Step 1: Velocity

s'(t) = 3t^2 - 12t + 9

.

2. Step 2: Acceleration

s''(t) = 6t - 12

, so

s''(2) = 12 - 12 = 0

.

Practice 02Hard

For the circle

x^2 + y^2 = 25

, find

\dfrac{d^2y}{dx^2}

in terms of

x

and

y

.

Need a hint?

Get

\frac{dy}{dx}

implicitly first, then differentiate that quotient with respect to

x

, substituting

\frac{dy}{dx} = -x/y

back in at the end.

Show solution

1. Step 1: First Derivative (Implicit)

2x + 2y\,\dfrac{dy}{dx} = 0 \implies \dfrac{dy}{dx} = -\dfrac{x}{y}

.

2. Step 2: Differentiate Again (Quotient Rule)

\dfrac{d^2y}{dx^2} = -\dfrac{(1)(y) - x\,\dfrac{dy}{dx}}{y^2} = -\dfrac{y - x\,\dfrac{dy}{dx}}{y^2}

.

3. Step 3: Substitute dy/dx Back In

\dfrac{d^2y}{dx^2} = -\dfrac{y - x(-x/y)}{y^2} = -\dfrac{y + x^2/y}{y^2} = -\dfrac{y^2 + x^2}{y^3} = -\dfrac{25}{y^3}

(using

x^2 + y^2 = 25

).

Common Pitfalls

⚠
Forgetting dy/dx Reappears When Differentiating ImplicitlyWhen you differentiate $-x/y$ with respect to $x$ , the $y$ in the denominator is a function of $x$ , so its derivative brings another $\dfrac{dy}{dx}$ into the expression — don't treat $y$ as a constant.
⚠
$f^{(4)}$ Is Not $f^4$ $f^{(4)}(x)$ (in parentheses) is the fourth derivative; $f(x)^4$ is the function raised to the fourth power. The parentheses matter.

Worked Examples

1. Step 1: Differentiate Both Sides

2. Step 2: Solve for dy/dx

3. Step 3: Evaluate the Slope at (3, -4)

4. Step 4: Write the Tangent Line

1. Step 1: Differentiate Both Sides

2. Step 2: Collect All y' Terms on One Side

3. Step 3: Solve for y'

Differentiate Every Term w.r.t. x

Collect and Factor

Substitute the Point Last

Practice — Your Turn

1. Step 1: Differentiate Once for y'

2. Step 2: Evaluate y' at (1, 0)

3. Step 3: Differentiate the y' Equation Again for y''

4. Step 4: Substitute x=1, y=0, y'=-2

1. Step 1: Find the Slope of y = ax³ in Terms of x, y

2. Step 2: Find the Slope of x² + 3y² = b Implicitly

3. Step 3: Multiply the Two Slopes

4. Step 4: Conclude

1. Step 1: Differentiate Implicitly

2. Step 2: Set the Numerator to Zero

3. Step 3: Substitute Back into the Original Equation

4. Step 4: State the Points

Worked Examples

1. Step 1: Differentiate the Outer e^(...) Layer

2. Step 2: Differentiate sec(3θ) with the Chain Rule

3. Step 3: Combine

1. Step 1: Rewrite as a Power

2. Step 2: Differentiate the Outer Power

3. Step 3: Multiply by the Inner Derivative

Classify the Expression

For Logarithmic Differentiation, Take ln of Both Sides First

Don't Forget the Chain Rule for Nested Arguments

Practice — Your Turn

1. Step 1: Take the Natural Log of Both Sides

2. Step 2: Differentiate Both Sides Implicitly

3. Step 3: Combine the Right Side

4. Step 4: Solve for y' by Multiplying by y

1. (a) Constant Base, Constant Exponent

2. (b) Variable Base, Constant Exponent → Power Rule

3. (c) Constant Base, Variable Exponent → Exponential Rule

4. (d) Variable Base AND Variable Exponent → Logarithmic Differentiation

1. Step 1: Set the Derivative Equal to the Target Slope

2. Step 2: Solve for x

3. Step 3: Find the y-Coordinate

4. Step 4: State the Point

Worked Examples

1. Step 1: First Derivative

2. Step 2: Second Derivative

3. Step 3: Third Derivative

1. Step 1: First Derivative

2. Step 2: Second Derivative

Differentiate, Then Differentiate the Result

For Implicit Second Derivatives, Substitute Last

Practice — Your Turn

1. Step 1: Velocity

2. Step 2: Acceleration

1. Step 1: First Derivative (Implicit)

2. Step 2: Differentiate Again (Quotient Rule)

3. Step 3: Substitute dy/dx Back In

Gary Chang