Question

If $y={{x}^{{{x}^{{{x}^{.....\infty }}}}}}$ , then $y'$ is equal to:
(a) $\dfrac{-{{y}^{2}}}{x\left( 1-y\log x \right)}$
(b) $\dfrac{{{y}^{2}}}{1-y\log x}$
(c) $\dfrac{{{y}^{2}}}{x\left( 1-y\log x \right)}$
(d) $\dfrac{-{{y}^{2}}}{1-y\log x}$

Answer 1

Hint: Start by using $y={{x}^{{{x}^{{{x}^{.....\infty }}}}}}$ and taking log of both the sides of the equation. Now use the identity that $\log {{a}^{b}}=b\log a$ , which gives $\log y={{x}^{{{x}^{{{x}^{.....\infty }}}}}}\log x$ , which comes to be equal to $\log y=y\log x$ . Now differentiate both sides of the equation with respect to x. Use the uv rule of differentiation, i.e., $\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$ .

Complete step-by-step solution -
The equation given in the question is:
$y={{x}^{{{x}^{{{x}^{.....\infty }}}}}}......(i)$
Now we know, if we take logs of both sides of the equation, we know that the equation remains valid. So, we will take logs of both sides of the equation. On doing so, we get
$\log y=\log {{x}^{{{x}^{{{x}^{.....\infty }}}}}}$
Now, we know that $\log {{a}^{b}}=b\log a$ . So, using this identity in our equation, we get
$\log y={{x}^{{{x}^{{{x}^{.....\infty }}}}}}\log x$
Now, we know that if we subtract 1 from infinity, we again get infinity. So, if we use equation (i), we get
 $\log y=y\log x$
Now we will differentiate both sides of the equation with respect to x. On doing so, we get
 $\dfrac{d\left( \log y \right)}{dx}=\dfrac{d\left( y\log x \right)}{dx}$
Using the uv rule of differentiation, i.e., $\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$ and $\dfrac{d\left( \log y \right)}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}$ , we get
$\dfrac{1}{y}\dfrac{dy}{dx}=\log x\dfrac{dy}{dx}+\dfrac{yd\left( \log x \right)}{dx}$
Now, we know that the derivative of $\log x\text{ is }\dfrac{1}{x}$ . So, using this in our equation, we get
$\dfrac{1}{y}\dfrac{dy}{dx}=\log x\times \dfrac{dy}{dx}+y\times \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}-\log x\times \dfrac{dy}{dx}=\dfrac{y}{x}$
Now, we will take $\dfrac{dy}{dx}$ common and solve. On doing so, we get
$\left( \dfrac{1}{y}-\log x \right)\dfrac{dy}{dx}=\dfrac{y}{x}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x\left( \dfrac{1}{y}-\log x \right)}$
Now we know that $\dfrac{dy}{dx}$ is y’. So, our equation becomes
$y'=\dfrac{{{y}^{2}}}{x\left( 1-y\log x \right)}$
Therefore, we can conclude that the answer to the above question is option (c).

Note: Be careful while reporting the answer and calculating it as there are two options with the same terms and differ just by the sign. Also, remember that for taking log of both the sides of the equation, both the sides must be positive. In the above question as nothing is mentioned so we consider both sides of the equation to be positive and take log on both sides.