AP Calculus AB and BC

Logarithmic vs Implicit Differentiation

Q: For x to the x, is that a power rule problem or a log differentiation problem?

Logarithmic differentiation. The power rule \frac{d}{dx}x^{n} = nx^{n-1} is stated for a constant n, and here the exponent moves with x, so the rule does not apply. Taking logs first gives y' = x^{x}\left(\ln x + 1\right).

Q: Why does y prime over y keep appearing?

Because y is a function of x, so differentiating \ln y needs the chain rule and produces \frac{1}{y}\cdot\frac{dy}{dx}. Multiplying both sides by y at the end is what puts the original function back into the answer.

Logarithmic differentiation is implicit differentiation applied to the equation ln y equals ln f of x, so it is a special case rather than a rival. Reach for it when the exponent contains x, since neither the power rule nor the exponential rule applies there.

Logarithmic differentiation

Use when: The variable appears in the exponent, or the function is a long product or quotient that logs would flatten into a sum of simple terms.

Implicit differentiation

Use when: You are handed an equation relating x and y that is not solved for y, and solving it would be ugly, impossible, or would break the function into branches.

Side by side

	Logarithmic differentiation	Implicit differentiation
Where the equation comes from	You build it, by writing $\ln y = \ln f(x)$	You are given it, as in $x^{2} + y^{2} = 25$
When to reach for it	The exponent contains $x$ , as in $y = x^{x}$	The relation cannot be solved for $y$
Opening move	Take $\ln$ of both sides and expand with log rules	Differentiate both sides with respect to $x$
Shape of the answer	$dy/dx$ in terms of $x$ alone, after substituting $y = f(x)$	$dy/dx$ usually in terms of both $x$ and $y$
Relationship between them	A special case, applied to an equation you constructed	The general technique that the other one runs on

The power rule wants a constant exponent and the exponential rule wants a constant base, so $y = x^{x}$ is outside the reach of both. Logs repair that in one step, since $\ln\left(x^{x}\right) = x \ln x$ pulls the exponent down into a product the product rule can differentiate. Take logs of both sides, differentiate, and solve for $y'$ .

y = x^{x} \;\Longrightarrow\; \ln y = x \ln x \;\Longrightarrow\; \frac{y'}{y} = \ln x + 1 \;\Longrightarrow\; y' = x^{x}\left(\ln x + 1\right)

The middle step is implicit differentiation and nothing else. Differentiating $\ln y$ with respect to $x$ gives $\frac{1}{y} \cdot \frac{dy}{dx}$ by the chain rule, exactly the move that turns $x^{2} + y^{2} = 25$ into $2x + 2y\frac{dy}{dx} = 0$ and then $\frac{dy}{dx} = -\frac{x}{y}$ . Only the origin of the equation differs. One detail worth keeping: $\ln y$ requires $y > 0$ , so the careful version starts from $\ln\lvert y \rvert$ , whose derivative is the same $\frac{y'}{y}$ .

The two wrong derivatives of x to the x

Forcing the power rule onto $x^{x}$ gives $x \cdot x^{x-1}$ , which collapses back to $x^{x}$ . Forcing the exponential rule onto it gives $x^{x}\ln x$ . Both are wrong, and the giveaway is that the correct answer $x^{x}\left(\ln x + 1\right)$ is their sum. Treat that as a check on your work, not as a rule to memorise: once $x$ sits in the base and the exponent at once, either take logs and differentiate implicitly or rewrite as $e^{x \ln x}$ and use the chain rule.

Frequently asked questions

For x to the x, is that a power rule problem or a log differentiation problem?

Logarithmic differentiation. The power rule $\frac{d}{dx}x^{n} = nx^{n-1}$ is stated for a constant $n$ , and here the exponent moves with $x$ , so the rule does not apply. Taking logs first gives $y' = x^{x}\left(\ln x + 1\right)$ .

Is logarithmic differentiation on the AP exam?

It is not a named technique in the Course and Exam Description, so a question that can only be done this way is unlikely. It is still worth having, because together with the rewrite $x^{\sin x} = e^{\sin x \ln x}$ it is one of only two routes to a variable base with a variable exponent, and it turns a four-factor quotient into a sum of four easy derivatives.

Why does y prime over y keep appearing?

Because $y$ is a function of $x$ , so differentiating $\ln y$ needs the chain rule and produces $\frac{1}{y}\cdot\frac{dy}{dx}$ . Multiplying both sides by $y$ at the end is what puts the original function back into the answer.

In the CED: Unit 3: Chain Rule, Implicit, and Inverses