Question

If ${e^y} (X + 1) = 1,$ then show that $\dfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\dfrac{{dy}}{{dx}}} \right)^2}.$

Answer 1

Hint: I this question, so we have to prove $\dfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$ to solve this type of question first we try to differentiate the given term with respect to $dx$ , and find out $\dfrac{{dy}}{{dx}}$ . After finding out the value of $\dfrac{{dy}}{{dx}}$ , double differentiate it and also do the square of $\dfrac{{dy}}{{dx}}$and find the value. How if the obtained value and square value of $\dfrac{{dy}}{{dx}}$ will equal then it will be proved.

Complete step-by-step solution:
Given: ${e^y} (X + 1) = 1,$ show that $\dfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\dfrac{{dy}}{{dx}}} \right)^2}.$
The given term will be differentiating with respect to $dx$ so, when we differentiate these terms following the chain rule; we have,
$\because {e^y} = (X + 1) = 1$
Now, differentiating the equate with respect to $dx$
$ \Rightarrow {e^y}\dfrac{{d(x + 1)}}{{dx}} + (x + 1)\dfrac{{d{e^y}}}{{dx}} = \dfrac{{d1}}{{dx}}$ (Using chain rule)
$
   \Rightarrow {e^y}\left( {\dfrac{{dx}}{{dx}} + \dfrac{{d1}}{{dx}}} \right) + (x + 1){e^y}\dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow {e^y}(1 + 0) + {e^y}(x + 1)\dfrac{{dy}}{{dx}} = 0 \\
  \therefore \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{(x + 1)}} \\
 $
Now, again differentiating with respect to $dx$
$
  \therefore \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{(x + 1)\dfrac{{d( - 1)}}{{dx}} - ( - 1)\dfrac{{d(x + 1)}}{{dx}}}}{{{{(x + 1)}^2}}} \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = + \dfrac{1}{{{{(x + 1)}^2}}} \\
  \therefore \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{{{(x + 1)}^2}}} \\
 $
Now, we have value of $\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{{{(x + 1)}^2}}}$ and we also find out the value of $\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{(x + 1)}}$
Now, squaring this equation on both sides; we have,
$\therefore {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \dfrac{1}{{{{(x + 1)}^2}}}$
Hence, ${\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \dfrac{{{d^2}y}}{{d{x^2}}}$ proved.

Note: In this question, we use the formula $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}}$ so a student must know about the basic formula of differentiating. So, it will help to solve these types of questions and one more thing always follow the chain rule of differentiation.