Question

If \(e^y(x+2)=10\), then \(\frac{d^2y}{dx^2}\) is equal to:

• A. \((\frac{dy}{dx})^2\)
• B. \((\frac{dy}{dx})^3\)
• C. \((\frac{dy}{dx})\)
• D. \(- (\frac{dy}{dx})^2\)

Answer 1

The Correct Option is A

We begin with the given equation:

\(e^y(x+2)=10\)

To find \(\frac{d^2y}{dx^2}\), we'll first differentiate both sides with respect to \(x\) using implicit differentiation.

Differentiate the left-hand side:

\(\frac{d}{dx}[e^y(x+2)] = e^y\frac{d}{dx}(x+2) + (x+2)\frac{d}{dx}(e^y)\)

\(= e^y \cdot 1 + (x+2)e^y \cdot \frac{dy}{dx}\)

\(= e^y + (x+2)e^y\frac{dy}{dx}\)

The derivative of the right-hand side (10) with respect to \(x\) is zero:

\(\frac{d}{dx}[10] = 0\)

Setting the derivatives equal gives:

\(e^y + (x+2)e^y\frac{dy}{dx} = 0\)

We solve for \(\frac{dy}{dx}\):

\((x+2)e^y\frac{dy}{dx} = -e^y\)

\(\frac{dy}{dx} = \frac{-e^y}{(x+2)e^y}\)

\(\frac{dy}{dx} = \frac{-1}{x+2}\)

Now, differentiate \(\frac{dy}{dx}\) with respect to \(x\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{-1}{x+2}\right)\)

Using the quotient rule or recognizing this as a simple derivative of \(-1\) times a reciprocal function:

\(\frac{d}{dx}\left(\frac{-1}{x+2}\right) = \frac{0(x+2) - (-1)}{(x+2)^2}\)

\(= \frac{1}{(x+2)^2}\)

Notice the sign, the negative remains inside:

\(\frac{d^2y}{dx^2} = \left(\frac{1}{(x+2)^2}\right) = \left(\frac{1}{x+2}\right)^2\)

Since \(\frac{dy}{dx} = \frac{-1}{x+2}\), we find:

\(\left(\frac{dy}{dx}\right)^2 = \left(\frac{-1}{x+2}\right)^2\)

\(= \left(\frac{1}{x+2}\right)^2\)

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