Q.Find $\frac{dy}{dx}$ in the following: $x^x - 2^{\sin x}$

Concept: Implicit Differentiation (using logarithmic differentiation for the variable-exponent term).

We have $y = x^x - 2^{\sin x}$. Differentiate term by term.

Step 1: For $u = x^x$, take $\log u = x \log x$. Differentiate: $\frac{1}{u} \frac{du}{dx} = \log x + 1$, so $\frac{du}{dx} = x^x (1 + \log x)$. …

We have $y = x^x - 2^{\sin x}$. The first term, $x^x$, is a classic case where the variable is both the base and the exponent — ordinary power rule or exponential rule alone won't work. The second term, $2^{\sin x}$, is an exponential function with a constant base but a variable exponent, so we can handle it with the exponential rule plus the chain rule.

The key technique for $x^x$ is logarithmic differentiation: take the natural log of both sides of $u = x^x$, differentiate implicitly, then solve for $u'$. This works because the logarithm converts the exponent into a product, which we can differentiate using the product rule.

Let's break it down step by step.

1. Set up the two parts separately.

   Let $u = x^x$ and $v = 2^{\sin x}$, so $y = u - v$. Then $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$.

2. Differentiate $u = x^x$ using logarithmic differentiation.

   Take the natural log of both sides:

Now differentiate both sides with respect to $x$. On the left, by the chain rule, $\frac{d}{dx} \log u = \frac{1}{u} \frac{du}{dx}$. On the right, use the product rule:

So we have:

Multiply both sides by $u$:

$\frac{du}{dx} = u (\log x + 1) = x^x (1 + \log x).$ …

Method: Differentiating a Sum of Two Different Exponential Forms

Use this method when the expression is a sum or difference of several terms, each of a different exponential "shape" — here $x^x$ (variable base and exponent) and $2^{\sin x}$ (constant base, variable exponent). Never try to take one single logarithm of the whole sum; logarithms don't distribute over addition or subtraction.

Steps

Step 1: Split the expression into separate terms and differentiate each independently.

Each term gets its own, term-appropriate rule — this is what keeps the algebra manageable.

Step 2: For the variable-base, variable-exponent term ($x^x$), use logarithmic differentiation.

Step 3: For the constant-base, variable-exponent term ($2^{\sin x}$), use the exponential rule combined with the Chain Rule. …

Common Mistakes

Mistake 1: Taking the logarithm of the entire expression $x^x - 2^{\sin x}$ at once.

Why it's wrong: $\log(A-B)$ is not $\log A - \log B$ — logarithms only simplify products, quotients, and powers, never sums or differences. Correct approach: split the expression into its separate additive terms first, and only take logs of an individual term if that term itself is a variable-base power.

Mistake 2: Forgetting the $\log 2$ factor when differentiating $2^{\sin x}$.

Why it's wrong: for a constant base $a\ne e$, $\frac{d}{dx}a^{f(x)} = a^{f(x)}\log a\cdot f'(x)$ — the $\log a$ factor is easy to drop, especially since it's absent when the base is $e$. Correct approach: always check whether the base is $e$ (no extra factor) or some other constant (multiply by $\log a$). …