Q.Find $\frac{dy}{dx}$ in the following: $(\sin x - \cos x)^{(\sin x - \cos x)}, \frac{\pi}{4} < x < \frac{3\pi}{4}$

Concept: Implicit Differentiation (using logarithmic differentiation for a variable exponent).

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

Step 1: Take the natural logarithm of both sides:

Step 2: Differentiate both sides with respect to $x$ (using the product rule on the right):

Step 3: Simplify the second term: $(\sin x - \cos x) \cdot \frac{\cos x + \sin x}{\sin x - \cos x} = \cos x + \sin x$. So:

$\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) \big[ \log(\sin x - \cos x) + 1 \big]$ …

We have $y = (\sin x - \cos x)^{(\sin x - \cos x)}$. This is a variable raised to a variable power — neither a pure power rule nor a pure exponential rule applies directly. The standard technique for such "function to the power of function" forms is logarithmic differentiation.

Why does this work? Taking the natural logarithm converts the exponent into a product, which we can then differentiate using the product rule. The chain rule then handles the composition on the left side.

Let’s go step by step.

1. Take the natural logarithm of both sides. Since $y > 0$ in the given interval $\frac{\pi}{4} < x < \frac{3\pi}{4}$ (check: $\sin x - \cos x$ is positive here), we can safely write:

Using the power property of logs:

2. Differentiate both sides with respect to $x$. On the left, by the chain rule:

On the right, we have a product: $u \cdot v$ where $u = \sin x - \cos x$ and $v = \log(\sin x - \cos x)$.

Differentiate using the product rule:

3. Find $u'$ and $v'$.

For $v = \log(\sin x - \cos x)$, use the chain rule:

4. Assemble the derivative of the right side.

The second term simplifies: $(\sin x - \cos x) \cdot \frac{\cos x + \sin x}{\sin x - \cos x} = \cos x + \sin x$.

So the right side becomes:

Factor out $(\cos x + \sin x)$:

5. Equate and solve for $\frac{dy}{dx}$. From step 2: …

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 this as a simple power rule or exponential rule because the base and exponent "look the same".

Why it's wrong: even though base and exponent are literally identical here, both still depend on $x$ — the function is genuinely $u^u$-shaped, which needs logarithmic differentiation regardless of the base and exponent happening to match. Correct approach: check for $x$-dependence in both positions, not whether they're equal to each other.

Mistake 2: Forgetting to verify the domain condition that makes $\sin x-\cos x>0$ (so the logarithm is defined). …