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

OR

If $x = a( heta - \sin heta)$, $y = a(1+\cos heta)$, then find $\dfrac{dy}{dx}$.

Question

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

OR

If $x = a(	heta - \sin	heta)$, $y = a(1+\cos	heta)$, then find $\dfrac{dy}{dx}$.

Accepted Answer

> [!TLDR]
> Differentiate $e^y(x+1)=1$ implicitly twice; the two results match term-by-term. (OR: parametric differentiation with $	heta$.)
>
> **Main part.** Given $e^y(x+1)=1$.
>
> Differentiate both sides w.r.t. $x$ (product rule):
> $$e^y\frac{dy}{dx}(x+1) + e^y\cdot 1 = 0 \;\Rightarrow\; e^y\left[(x+1)\frac{dy}{dx}+1ight]=0$$
> Since $e^y
e 0$:
> $$(x+1)\frac{dy}{dx} = -1 \;\Rightarrow\; \frac{dy}{dx} = -\frac{1}{x+1}$$
> Differentiate again w.r.t. $x$:
> $$\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2}$$
> Also,
> $$\left(\frac{dy}{dx}ight)^2 = \left(-\frac{1}{x+1}ight)^2 = \frac{1}{(x+1)^2}$$
> Hence $\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}ight)^2$. Hence Proved.
>
> **OR.** Given $x=a(	heta-\sin	heta)$, $y=a(1+\cos	heta)$.
>
> $$\frac{dx}{d	heta} = a(1-\cos	heta), \qquad \frac{dy}{d	heta} = -a\sin	heta$$ …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

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

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