Q.Find $\frac{dy}{dx}$ in the following: $x^{\sin x} + (\sin x)^{\cos x}$

The key idea is Implicit Differentiation combined with logarithmic differentiation, since the variable appears in both the base and the exponent.

Let $y = x^{\sin x} + (\sin x)^{\cos x}$.

Set $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$, so $y = u + v$.

Step 1: For $u = x^{\sin x}$, take $\log u = \sin x \cdot \log x$. Differentiate:

Thus

Step 2: For $v = (\sin x)^{\cos x}$, take $\log v = \cos x \cdot \log(\sin x)$. Differentiate:

Thus

$\frac{dv}{dx} = (\sin x)^{\cos x} \left( -\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x} \right)$ …

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The problem asks for $\frac{dy}{dx}$ where $y = x^{\sin x} + (\sin x)^{\cos x}$. At first glance, this looks like a sum of two power functions, but the exponents are not constants — they are functions of $x$. That means neither the standard power rule ($\frac{d}{dx} x^n = n x^{n-1}$) nor the exponential rule ($\frac{d}{dx} a^x = a^x \log a$) applies directly. Each term has the variable in both the base and the exponent.

The technique that handles this is logarithmic differentiation — or equivalently, rewriting each term as $e^{\text{something}}$ and then differentiating. The idea is simple: for any positive base $f(x)$ and any real exponent $g(x)$, we have $f(x)^{g(x)} = e^{g(x) \log f(x)}$. This turns the problem into differentiating a composition of functions, which we can handle with the chain rule and product rule.

Let’s do it step by step.

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1. Rewrite the function Since both $x$ and $\sin x$ are positive for the relevant domain (we assume $x>0$ and $\sin x > 0$ for the expression to be real), we can write:

2. Differentiate term by term

   Let $u = e^{\sin x \log x}$ and $v = e^{\cos x \log(\sin x)}$, so $y = u + v$ and $\frac{dy}{dx} = u' + v'$.

3. Find $u'$

   For $u = e^{\sin x \log x}$, the derivative is:

The factor $e^{\sin x \log x}$ is just $x^{\sin x}$, so:

Now differentiate $\sin x \log x$ using the product rule:

Therefore:

4. Find $v'$ For $v = e^{\cos x \log(\sin x)}$, we have:

The factor $e^{\cos x \log(\sin x)}$ is $(\sin x)^{\cos x}$, so:

Differentiate $\cos x \log(\sin x)$ using the product rule:

Simplify the second term: $\cos x \cdot \frac{\cos x}{\sin x} = \frac{\cos^2 x}{\sin x}$.

So:

Therefore: …

Method: Logarithmic Differentiation Applied Separately to Each Term of a Sum

Use this method when $y$ is a sum of two terms, and both terms individually have a variable base raised to a variable exponent — here $x^{\sin x}$ and $(\sin x)^{\cos x}$. Even though both terms need the same technique, they must still be handled completely separately because the sum itself cannot be log-differentiated as one piece.

Steps

Step 1: Split the sum into two labelled terms.

Step 2: Log-differentiate $u$.

(Product Rule on the right, since $\sin x$ and $\log x$ are both functions of $x$ multiplied together.)

Step 3: Log-differentiate $v$, being careful with the Chain Rule inside $\log(\sin x)$.

$\log v = \cos x\cdot\log(\sin x) \;\Rightarrow\; \frac{1}{v}\frac{dv}{dx} = -\sin x\log(\sin x) + \cos x\cdot\frac{\cos x}{\sin x}$ …

Common Mistakes

Mistake 1: Applying the power rule to $x^{\sin x}$ as if $\sin x$ were a fixed number.

Why it's wrong: writing $\sin x\cdot x^{\sin x - 1}$ treats the exponent as constant and completely ignores that $\sin x$ itself changes with $x$ — this misses the $\cos x\log x$ term that comes from differentiating the exponent. Correct approach: whenever the exponent is a function of $x$ (not a fixed number), logarithmic differentiation is required, not the ordinary power rule.

Mistake 2: Losing the Chain Rule factor when differentiating $\log(\sin x)$ inside the second term.

Why it's wrong: $\frac{d}{dx}\log(\sin x) = \frac{\cos x}{\sin x} = \cot x$ — some students write just $\frac{1}{\sin x}$ and forget to multiply by $\cos x$, the derivative of $\sin x$. Correct approach: always identify $\sin x$ as the inner function of the log and multiply by its own derivative. …