Calculus strategy — differentiation techniques decision tree

Question

Differentiate $y = x^x$ with respect to $x$ .

(JEE Main 2023 pattern — requires logarithmic differentiation)

Differentiation Technique Selection

flowchart TD
    A["Function to Differentiate"] --> B{What form?}
    B -->|"x^n, sin x, e^x, log x"| C["Standard formulas"]
    B -->|"f(g(x)) — function of function"| D["Chain Rule"]
    B -->|"u(x) * v(x)"| E["Product Rule"]
    B -->|"u(x) / v(x)"| F["Quotient Rule"]
    B -->|"x^x or f(x)^g(x)"| G["Logarithmic Differentiation"]
    B -->|"x = f(t), y = g(t)"| H["Parametric: dy/dx = (dy/dt)/(dx/dt)"]
    B -->|"f(x,y) = 0 (implicit)"| I["Implicit Differentiation"]
    G --> G1["Take ln both sides, then differentiate"]
    D --> D1["dy/dx = f'(g(x)) * g'(x)"]

Solution — Step by Step

$y = x^x$ has both the base and exponent as variables. We cannot use the power rule ( $d/dx[x^n] = nx^{n-1}$ , where $n$ is constant) or the exponential rule ( $d/dx[a^x] = a^x \ln a$ , where $a$ is constant). When both vary, we need logarithmic differentiation.

\ln y = \ln(x^x) = x\ln x

Now differentiate both sides with respect to $x$ :

\frac{1}{y}\cdot\frac{dy}{dx} = \frac{d}{dx}(x\ln x)

Apply the product rule to the right side:

\frac{1}{y}\cdot\frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1

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

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

Why This Works

Logarithmic differentiation converts products into sums and powers into products, making complex expressions manageable. Taking $\ln$ of both sides brings the exponent down as a coefficient. Then implicit differentiation (the $1/y \cdot dy/dx$ on the left) handles the rest. This technique works for any $f(x)^{g(x)}$ form.

Alternative Method — Rewrite as Exponential

Since $x^x = e^{x\ln x}$ , we can differentiate using the chain rule:

\frac{d}{dx}[e^{x\ln x}] = e^{x\ln x} \cdot \frac{d}{dx}(x\ln x) = x^x(\ln x + 1)

Same answer, slightly different pathway. Some students find this approach more intuitive.

Logarithmic differentiation is also useful for products of many functions. To differentiate $y = \frac{(x+1)^2(x+2)^3}{(x+3)^4}$ , take $\ln$ of both sides to convert the product/quotient into a sum. This avoids nested product and quotient rules entirely.

Common Mistake

When differentiating $\ln y$ with respect to $x$ , students write $1/y$ instead of $(1/y)(dy/dx)$ . The chain rule is essential here — $y$ is a function of $x$ , not $x$ itself. Forgetting the $dy/dx$ factor means the left side is wrong, and the entire answer becomes incorrect despite correct work on the right side.