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# If ${\text{y = }}{{\text{x}}^{\text{x}}}$ , prove that $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$

Last updated date: 08th Aug 2024
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Hint: First we apply the logarithm function on the both sides of the equation ${\text{y = }}{{\text{x}}^{\text{x}}}$, then we’ll proceed towards finding the first derivative of the equation with respect to the independent variable i.e. x. similarly, we’ll find the double derivative of the function. After getting the first and second derivates of the function we’ll use the substitution method to eliminate the terms that are not required to get our answer.

Given data: ${\text{y = }}{{\text{x}}^{\text{x}}}$
On applying logarithm function on both sides, we get
${\text{ln(y) = ln(}}{{\text{x}}^{\text{x}}}{\text{)}}$
It is well known that, ${\text{ln}}{{\text{a}}^{\text{b}}}{\text{ = blna}}$,
$\Rightarrow {\text{lny = xln(x)}}$
Now, differentiating both sides with respect to x,
Using, chain rule and multiplication rule i.e.
$\dfrac{{{\text{df(z)}}}}{{{\text{dx}}}}{\text{ = f'(z)}}\dfrac{{{\text{dz}}}}{{{\text{dx}}}}{\text{, and}} \\ {\text{d(uv) = udv + vdu}} \\$
$\dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = x}}\left( {\dfrac{{\text{1}}}{{\text{x}}}} \right){\text{ + lnx}} \\ \Rightarrow \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 1 + lnx}}..........{\text{(i)}} \\$
Multiplying ‘y’ with the whole equation
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = y + ylnx}}................{\text{(ii)}}$
Again, on differentiating both sides with respect to x,
$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ + }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{lnx}} \\ \Rightarrow \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(1 + lnx) + }}\dfrac{{\text{y}}}{{\text{x}}}..........{\text{(iii)}} \\$
Now, substituting the value of (1+lnx) from equation(i) to equation(iii), we’ll be left with
$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(}}\dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{) + }}\dfrac{{\text{y}}}{{\text{x}}} \\ \Rightarrow \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ + }}\dfrac{{\text{y}}}{{\text{x}}} \\ \Rightarrow \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}} \\$
Hence, we obtained our equation i.e. $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$

Note: Alternative solution for this question can be done by substituting the value of first and second derivatives directly to the equation that has to be proved.
$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$
On substituting the value of $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ and }}\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}$ to the left-hand side of the equation, we get,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(}}\dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{) + }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(y + ylnx)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}} \\ {\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{{{\text{y}}^{\text{2}}}}}{{\text{y}}}{{\text{(1 + lnx)}}^{\text{2}}} \\$
On simplification we get,
${\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
On substituting the value of $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$, we have
${\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(y + ylnx)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
On taking ${{\text{y}}^{\text{2}}}$out of the first term, we get,
${\text{ = }}\dfrac{{{{\text{y}}^{\text{2}}}}}{{\text{y}}}{{\text{(1 + lnx)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
${\text{ = y(1 + lnx}}{{\text{)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
${\text{ = 0}}$
i.e. equal to the right-hand side
Therefore, $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$, holds of ${\text{y = }}{{\text{x}}^{\text{x}}}$