Question

If $y=x^2+x^{\log x}$, then $\frac{dy}{dx}=?$

• A. $\frac{\left(x^2+\log x{x}^{\log x}\right)}{x}$
• B. $2\frac{\left(x^2+\log x{x}^{\log x}\right)}{x}$
• C. $\left(x^2+\log x{x}^{\log x}\right)$
• D. none of the above

Answer 1

The correct option is B

$2\frac{\left(x^2+\log x{x}^{\log x}\right)}{x}$

Explanation for the correct option:

Find the value of $\frac{dy}{dx}$:

Given,

$y=x^2+x^{\log x}$

Take $x^{\log x}=t...\left(1\right)$

Then,

$$\begin{gathered} \log t=\log x^{\log x} \\ \log t=\log x\log x\left[\because log{a}^b=bloga\right] \end{gathered}$$

Now differentiate with respect to $x$.

So,

$\begin{array}{rcl}\frac{1}{t}\frac{dt}{dx}& =& 2\log x\left(\frac{1}{x}\right)\\ \frac{dt}{dx}& =& 2t\log x\left(\frac{1}{x}\right)\\ \frac{dt}{dx}& =& \frac{2}{x}{x}^{\log x}\log x...\left(2\right)\left[\mathrm{from}1\right]\end{array}$

Now we can write given equation as

$y=x^2+t$

Now, differentiate the above equation with respect to $x$.

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& 2x+\frac{dt}{dx}\\ & =& 2x+\frac{2}{x}\log x{x}^{\log x}\mathbf{}\mathbf{}\mathbf{[}\mathbf{}\mathbf{from}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{]}\\ & =& 2\frac{\left(x^2+\log x{x}^{\log x}\right)}{x}\end{array}$

Hence, the correct option is B.