Question

Differentiate the following function with respect to x :

e^{x log x}

Answer 1

Let y = e^{x log x}

Taking log both the sides:

⇒ log y = log^{(e)x log x}

⇒ log y = x log x loge {log x^a = a log x}

⇒ log y = x log x {log e = 1}

Differentiating with respect to x:

Answer 2

Let \(x\log x \equiv u\)

By the chain rule if:

\(y = e^{x\log x}=e^u\)

Then

\(\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\frac{d}{du}(e^u)\cdot\frac{d}{dx}u\)

\(=e^{x\log x} \cdot \frac{d}{dx}(x\log x)\)

\(= e^{x\log x} \cdot \frac{d}{dx}(x)\log x + x\frac{d}{dx}(\log x)\)

Assuming by log x you mean natural logarithm (ln x):

\(\frac{dy}{dx}= e^{x\log x}\cdot\log x + 1\)