If $e^y(x + 1) = 1$ , which of these is true?

\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^2

Correct answerYour answer

\frac{d^2y}{dx^2} = \frac{dy}{dx}

Correct answerYour answer

\frac{d^2y}{dx^2} = e^y

Correct answerYour answer

\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$e^y = \frac{1}{x + 1}$ , so $y = -\log(x + 1)$ and $\frac{dy}{dx} = -\frac{1}{x + 1}$ .
$\frac{d^2y}{dx^2} = \frac{1}{(x + 1)^2}$ , which is $\left(-\frac{1}{x + 1}\right)^2 = \left(\frac{dy}{dx}\right)^2$ .

Worth noting

The minus sign belongs to the first derivative, $-\frac{1}{x + 1}$ , and squaring removes it.

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.

If ey(x+1)=1e^y(x + 1) = 1ey(x+1)=1, which of these is true?