Q.Find $\frac{dy}{dx}$ in the following: $\left(x+\frac{1}{x}\right)^x + x^{\left(1+\frac{1}{x}\right)}$

Write $y=u+v$ with $u=\left(x+\dfrac{1}{x}\right)^x$ and $v=x^{\,1+\frac{1}{x}}$, and use logarithmic differentiation on each (the variable is in both base and exponent).

Term $u$: $\log u=x\log\!\left(x+\tfrac{1}{x}\right)$, so

Term $v$: $\log v=\left(1+\tfrac{1}{x}\right)\log x$, so …

Each term has the variable in both base and exponent, so neither the power rule nor the exponential rule alone works — take logarithms. Write $y=u+v$ with

Differentiate $u$

Differentiate (product rule on the right):

Here $\dfrac{d}{dx}\!\left(x+\frac{1}{x}\right)=1-\dfrac{1}{x^2}=\dfrac{x^2-1}{x^2}$ and $x+\dfrac{1}{x}=\dfrac{x^2+1}{x}$, so

Therefore

Differentiate $v$

Product rule, with $\dfrac{d}{dx}\!\left(1+\frac{1}{x}\right)=-\dfrac{1}{x^2}$: …

Method: Term-by-Term Logarithmic Differentiation for a Sum of Two Variable-Exponent Expressions

Use this method when $y$ is a sum of two (or more) separate terms, and each term individually has the variable in both its base and its exponent — here $\left(x+\frac1x\right)^x$ and $x^{1+\frac1x}$. As with any sum, never take a single logarithm of the whole expression; log-differentiate each term on its own, then add the results.

Steps

Step 1: Split the sum and label each term.

Step 2: Logarithmically differentiate $u$ — take logs, then apply the Product Rule.

Differentiating the right side needs the Product Rule, and differentiating $\log\!\left(x+\frac1x\right)$ needs the Chain Rule (its inner derivative is $1-\frac{1}{x^2}$):

$\frac{1}{u}\frac{du}{dx} = \log\!\left(x+\frac1x\right) + x\cdot\frac{1-\frac1{x^2}}{x+\frac1x}$ …

Common Mistakes

Mistake 1: Attempting to take one logarithm of the entire sum $u+v$.

Why it's wrong: $\log(u+v)$ has no simplification in terms of $\log u$ and $\log v$, so this approach leads nowhere. Correct approach: always log-differentiate each additive term completely on its own before adding the two derivatives.

Mistake 2: Missing the Chain Rule when differentiating $\log\!\left(x+\frac1x\right)$.

Why it's wrong: the inner function $x+\frac1x$ has its own derivative, $1-\frac{1}{x^2}$, which must multiply the outer $\frac{1}{x+1/x}$ — simply writing $\frac{1}{x+1/x}$ alone skips this factor. Correct approach: treat $x+\frac1x$ as an inner function every time, and differentiate it explicitly before combining.

Mistake 3: Algebra errors when simplifying compound fractions like $\dfrac{1-1/x^2}{x+1/x}$. …