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

If ey(X+1)=1,{e^y} (X + 1) = 1, then show that d2ydx2=(dydx)2.\dfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\dfrac{{dy}}{{dx}}} \right)^2}.

Explanation

Solution

I this question, so we have to prove d2ydx2=(dydx)2\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 dxdx , and find out dydx\dfrac{{dy}}{{dx}} . After finding out the value of dydx\dfrac{{dy}}{{dx}} , double differentiate it and also do the square of dydx\dfrac{{dy}}{{dx}}and find the value. How if the obtained value and square value of dydx\dfrac{{dy}}{{dx}} will equal then it will be proved.

Complete step-by-step solution:
Given: ey(X+1)=1,{e^y} (X + 1) = 1, show that d2ydx2=(dydx)2.\dfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\dfrac{{dy}}{{dx}}} \right)^2}.
The given term will be differentiating with respect to dxdx so, when we differentiate these terms following the chain rule; we have,
∵ey=(X+1)=1\because {e^y} = (X + 1) = 1
Now, differentiating the equate with respect to dxdx
⇒eyd(x+1)dx+(x+1)deydx=d1dx\Rightarrow {e^y}\dfrac{{d(x + 1)}}{{dx}} + (x + 1)\dfrac{{d{e^y}}}{{dx}} = \dfrac{{d1}}{{dx}} (Using chain rule)
⇒ey(dxdx+d1dx)+(x+1)eydydx=0 ⇒ey(1+0)+ey(x+1)dydx=0 ∴dydx=−1(x+1)  \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 dxdx
∴d2ydx2=(x+1)d(−1)dx−(−1)d(x+1)dx(x+1)2 ⇒d2ydx2=+1(x+1)2 ∴d2ydx2=1(x+1)2  \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 d2ydx2=1(x+1)2\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{{{(x + 1)}^2}}} and we also find out the value of dydx=−1(x+1)\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{(x + 1)}}
Now, squaring this equation on both sides; we have,
∴(dydx)2=1(x+1)2\therefore {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \dfrac{1}{{{{(x + 1)}^2}}}
Hence, (dydx)2=d2ydx2{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \dfrac{{{d^2}y}}{{d{x^2}}} proved.

Note: In this question, we use the formula d(uv)dx\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.