Question

If $\mathrm{y}={\mathrm{x}}^{{\mathrm{x}}^{\mathrm{x}....\infty }},$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• A. ${\mathrm{yx}}^{\mathrm{y}-1}$
• B. $\frac{{\mathrm{y}}^2}{\mathrm{x}(1-\mathrm{ylog}\mathrm{x})}$
• C. $\frac{\mathrm{y}}{\mathrm{x}\left(1+\mathrm{ylogx}\right)}$
• D. None of these

Answer 1

The correct option is B

$\frac{{\mathrm{y}}^2}{\mathrm{x}(1-\mathrm{ylog}\mathrm{x})}$

Explanation for the correct option:

Finding the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$:

The given differential equation is $\mathrm{y}={\mathrm{x}}^{{\mathrm{x}}^{\mathrm{x}....\infty }}$

Let $y=x^{x....\infty }$,then

$\mathrm{y}={\mathrm{x}}^{\mathrm{y}}$

Taking $\log$ on both sides,

$\begin{array}{l}\mathrm{logy}=\mathrm{ylogx}\end{array}$

Differentiate the above equation with respect to $x$.

$\begin{array}{l}\left(\frac{1}{\mathrm{y}}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{y}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{logx}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\mathbf{logx}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{,}\mathbf{}\frac{\mathbf{d}\mathbf{(}\mathbf{a}\mathbf{·}\mathbf{b}\mathbf{)}}{\mathbf{dx}}\mathbf{=}\mathbf{a}\frac{\mathbf{db}}{\mathbf{dx}}\mathbf{+}\mathbf{b}\frac{\mathbf{da}}{\mathbf{dx}}\mathbf{]}\\ \Rightarrow \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\left(\frac{1}{\mathrm{y}}-\mathrm{logx}\right)=\frac{\mathrm{y}}{\mathrm{x}}\\ \Rightarrow \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\frac{\left(\frac{1}{\mathrm{y}}-\mathrm{logx}\right)}{\mathrm{y}}=\frac{\mathrm{y}}{\mathrm{x}}\\ \Rightarrow \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{{\mathrm{y}}^2}{\mathrm{x}\left(1-\mathrm{ylogx}\right)}\end{array}$

Hence, the correct option is(B).