Q.Differentiate the function $(x\cos x)^x + (x\sin x)^{\frac{1}{x}}$ with respect to $x$.

Let

$y = (x\cos x)^x + (x\sin x)^{\frac{1}{x}}$.

We differentiate each term separately using implicit differentiation (logarithmic differentiation).

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Term 1: $u = (x\cos x)^x$

Take $\log u = x \log(x\cos x)$.

Differentiate:

$\frac{1}{u} \frac{du}{dx} = \log(x\cos x) + x \cdot \frac{1}{x\cos x} \cdot (\cos x - x\sin x)$

Simplify:

$\frac{du}{dx} = (x\cos x)^x \left[ \log(x\cos x) + \frac{\cos x - x\sin x}{\cos x} \right]$.

Term 2: $v = (x\sin x)^{\frac{1}{x}}$

Take $\log v = \frac{1}{x} \log(x\sin x)$.

Differentiate:

$\frac{1}{v} \frac{dv}{dx} = -\frac{1}{x^2} \log(x\sin x) + \frac{1}{x} \cdot \frac{1}{x\sin x} \cdot (\sin x + x\cos x)$

Simplify: …

When a function has the variable in both the base and the exponent — like $u(x)^{v(x)}$ — the standard power rule or exponential rule alone won't work. The trick is to take natural logs first, which brings the exponent down as a multiplier, then differentiate implicitly. That's the heart of logarithmic differentiation.

Here we have a sum of two such terms:

Let’s call them $y_1$ and $y_2$:

Then $y' = y_1' + y_2'$. We'll handle each separately.

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1. Differentiate $y_1 = (x\cos x)^x$

Take natural log of both sides:

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

On the right, use the product rule:

Now differentiate $\log(x\cos x)$. Let $u = x\cos x$. Then

Differentiate $x\cos x$ using the product rule:

So

Putting it back:

Thus

Multiply through by $y_1$:

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2. Differentiate $y_2 = (x\sin x)^{1/x}$

Again, take logs:

Differentiate:

Use the product rule (or quotient rule — here it's a product of $x^{-1}$ and $\log(x\sin x)$):

Now differentiate $\log(x\sin x)$. Let $v = x\sin x$. Then

Differentiate $x\sin x$:

So …

Method: Logarithmic Differentiation for a Sum of Two Variable-Exponent Terms

Whenever you see two or more terms of the form $[\text{base}(x)]^{[\text{exponent}(x)]}$ added together, apply logarithmic differentiation to each term on its own — never try to combine them under one logarithm, since $\log(A+B) \ne \log A + \log B$.

Steps

Step 1: Isolate one term and take its own log

For a single term $u = f(x)^{g(x)}$, write

Differentiate treating this as a product $g(x)\cdot\log f(x)$.

Step 2: Differentiate the inner logarithm carefully

$\log f(x)$ itself may hide a product, e.g. $f(x)=x\cos x$. Use

and find $f'(x)$ by whatever rule $f$ needs (product rule if $f$ is itself a product).

Step 3: Solve for $\dfrac{du}{dx}$ and repeat for the second term …

Common Mistakes

Mistake 1: Taking one logarithm of the whole sum instead of one per term

Why it's wrong: $\log(A+B) \neq \log A + \log B$, so trying to logarithm the entire expression $(x\cos x)^x + (x\sin x)^{1/x}$ at once destroys the structure that makes logarithmic differentiation work. Correct approach: apply $\log$ separately to each $[\text{base}]^{[\text{exponent}]}$ term, differentiate each one on its own, and only add the two resulting derivatives at the very end.

Mistake 2: Dropping the negative sign when differentiating the $\frac{1}{x}$ exponent …