Day 8

Implicit And Exponential Logarithmic Derivatives

Implicit Differentiation Derivatives of Exponential & Logarithmic Functions

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).

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 Exercises

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)

NEED A HINT?

Differentiate both sides with respect to

x

; the

y^2

term needs the chain rule.

SHOW DETAILED EXPLANATION

Step 1: Differentiate Both Sides

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

Step 2: Solve for dy/dx

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

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

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

Step 4: Write the Tangent Line

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

Example 02Medium

Find

y'

for the folium of Descartes:

x^3+y^3=6xy

NEED A HINT?

The right side

6xy

needs the product rule, and every

y

term needs an extra

y'

tacked on.

SHOW DETAILED EXPLANATION

Step 1: Differentiate Both Sides

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

(the right side used the product rule on

6xy

Step 2: Collect All y' Terms on One Side

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

Step 3: Solve for y'

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

Example 03Hard

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 DETAILED EXPLANATION

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}

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

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

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]

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

Example 04AB/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 DETAILED EXPLANATION

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}

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

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

Step 3: Multiply the Two Slopes

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

Step 4: Conclude

Since the product of the slopes is

-1

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

Example 05Hard

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 DETAILED EXPLANATION

Step 1: Differentiate Implicitly

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

Step 2: Set the Numerator to Zero

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

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

Step 4: State the Points

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→

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 Exercises

Example 01Medium

Differentiate

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

NEED A HINT?

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.

SHOW DETAILED EXPLANATION

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]

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

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

Step 3: Combine

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

Example 02Medium

Differentiate

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

NEED A HINT?

Rewrite as

(\ln x)^{1/2}

and apply the chain rule, recalling that the derivative of

\ln x

\frac{1}{x}

SHOW DETAILED EXPLANATION

Step 1: Rewrite as a Power

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

Step 2: Differentiate the Outer Power

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

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}}

Example 03AB/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 DETAILED EXPLANATION

Step 1: Take the Natural Log of Both Sides

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

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}}

Step 3: Combine the Right Side

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

Step 4: Solve for y' by Multiplying by y

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

Example 04Medium

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 DETAILED EXPLANATION

(a) Constant Base, Constant Exponent

5^3

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

0

(b) Variable Base, Constant Exponent → Power Rule

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

(c) Constant Base, Variable Exponent → Exponential Rule

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

(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.

Example 05Easy

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 DETAILED EXPLANATION

Step 1: Set the Derivative Equal to the Target Slope

y' = e^x = 2

Step 2: Solve for x

x = \ln 2

Step 3: Find the y-Coordinate

y = e^{\ln 2} = 2

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.

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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Differentiate Every Term w.r.t. x

Collect and Factor

Substitute the Point Last

Practice Exercises

Step 1: Differentiate Both Sides

Step 2: Solve for dy/dx

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

Step 4: Write the Tangent Line

Step 1: Differentiate Both Sides

Step 2: Collect All y' Terms on One Side

Step 3: Solve for y'

Step 1: Differentiate Once for y'

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

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

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

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

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

Step 3: Multiply the Two Slopes

Step 4: Conclude

Step 1: Differentiate Implicitly

Step 2: Set the Numerator to Zero

Step 3: Substitute Back into the Original Equation

Step 4: State the Points

Classify the Expression

For Logarithmic Differentiation, Take ln of Both Sides First

Don't Forget the Chain Rule for Nested Arguments

Practice Exercises

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

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

Step 3: Combine

Step 1: Rewrite as a Power

Step 2: Differentiate the Outer Power

Step 3: Multiply by the Inner Derivative

Step 1: Take the Natural Log of Both Sides

Step 2: Differentiate Both Sides Implicitly

Step 3: Combine the Right Side

Step 4: Solve for y' by Multiplying by y

(a) Constant Base, Constant Exponent

(b) Variable Base, Constant Exponent → Power Rule

(c) Constant Base, Variable Exponent → Exponential Rule

(d) Variable Base AND Variable Exponent → Logarithmic Differentiation

Step 1: Set the Derivative Equal to the Target Slope

Step 2: Solve for x

Step 3: Find the y-Coordinate

Step 4: State the Point

Gary Chang