Q.Find $\frac{dy}{dx}$ in the following: $x^{x-3} + (x-3)^x$, for $x > 3$

Concept: Implicit Differentiation — but here we have two variable-base, variable-exponent terms, so we use logarithmic differentiation on each separately.

Let $y = x^{x-3} + (x-3)^x$. Differentiate term-by-term.

For $u = x^{x-3}$:

$\log u = (x-3)\log x$

$\frac{u'}{u} = \log x + \frac{x-3}{x}$

$u' = x^{x-3}\left(\log x + 1 - \frac{3}{x}\right)$

For $v = (x-3)^x$:

$\log v = x\log(x-3)$

$\frac{v'}{v} = \log(x-3) + \frac{x}{x-3}$

$v' = (x-3)^x\left(\log(x-3) + \frac{x}{x-3}\right)$

Add them: …

When you see a function where the variable appears in both the base and the exponent, the standard power rule or exponential rule alone won't work. For example, $x^n$ uses the power rule, and $a^x$ uses the exponential rule — but $x^{x-3}$ is neither. The technique that handles this is logarithmic differentiation: take the natural log of both sides, use log properties to bring the exponent down, then differentiate implicitly.

Here we have a sum of two such terms: $y = x^{x-3} + (x-3)^x$. Since the derivative of a sum is the sum of the derivatives, we can handle each term separately.

Let’s set $u = x^{x-3}$ and $v = (x-3)^x$, so $y = u + v$ and $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.

1. Differentiate $u = x^{x-3}$ Take $\log$ of both sides: $\log u = (x-3) \log x$. Differentiate implicitly with respect to $x$:

(using the product rule on the right).

So $\frac{du}{dx} = u \left( \log x + \frac{x-3}{x} \right) = x^{x-3} \left( \log x + \frac{x-3}{x} \right)$.

2. Differentiate $v = (x-3)^x$ Take $\log$: $\log v = x \log(x-3)$. Differentiate:

(again product rule).

So $\frac{dv}{dx} = v \left( \log(x-3) + \frac{x}{x-3} \right) = (x-3)^x \left( \log(x-3) + \frac{x}{x-3} \right)$.

3. Add the results …

Method: Logarithmic Differentiation for $y=[f(x)]^{g(x)}$

When BOTH the base and the exponent of a power contain $x$, neither the power rule (needs constant exponent) nor the exponential rule (needs constant base) applies directly. Logarithmic differentiation converts the exponent into a product, which can then be handled with the ordinary rules.

Steps

Step 1: Take the natural log of both sides

using the power property of logs, $\log(a^b)=b\log a$.

Step 2: Differentiate both sides with respect to $x$

On the left, $\frac{d}{dx}\log y=\frac{1}{y}\frac{dy}{dx}$ (chain rule, since $y$ is a function of $x$). On the right, apply the product rule (since it's now $g(x)$ times $\log f(x)$), with the chain rule on $\log f(x)$ itself.

Step 3: Solve for $\dfrac{dy}{dx}$ by multiplying both sides by $y$ …

Common Mistakes

Mistake 1: Treating $x^{x-3}$ as a plain power rule problem because the exponent "looks like it's just shifted by a constant".

Why it's wrong: $x-3$ still depends on $x$ — the exponent is not a fixed number, so the power rule (which needs a genuinely constant exponent) does not apply. Correct approach: check whether the exponent varies with $x$ at all, regardless of how simple it looks.

Mistake 2: Forgetting to differentiate the two terms with logarithmic differentiation independently, and instead trying to combine them into one log expression. …