Question

If $e^{x+y}+y=3x$, then $\dfrac{dy}{dx}$ is

• A. $\dfrac{3}{e^{x+y}+1}$
• B. $\dfrac{1}{e^{x+y}}$
• C. $\dfrac{1-e^{x+y}}{e^{x+y}}$
• D. $\dfrac{3-e^{x+y}}{e^{x+y}+1}$

Hint:

When differentiating expressions like \(e^{x+y}\), always apply the \textbf{chain rule}. Since \(y\) depends on \(x\), the derivative becomes \(e^{x+y}(1+\frac{dy}{dx})\).

Answer 1

The Correct Option is D

To find the derivative \(\frac{dy}{dx}\) from the given equation, we will use implicit differentiation. The given equation is:

\(e^{x+y} + y = 3x\)

We need to differentiate both sides of the equation with respect to \(x\). Remember that \(y\) is also a function of \(x\), so apply the chain rule when differentiating terms involving \(y\).

The derivative of \(e^{x+y}\) with respect to \(x\) is \(e^{x+y} \times (1 + \frac{dy}{dx})\). This is because the derivative of \(e^u\) where \(u = x + y\) is \(e^{x+y}\), and applying the chain rule gives us a factor of \((1 + \frac{dy}{dx})\).

The derivative of \(y\) with respect to \(x\) is simply \(\frac{dy}{dx}\).

The derivative of \(3x\) with respect to \(x\) is \(3\).

Putting all these together, we have:

\(e^{x+y}(1 + \frac{dy}{dx}) + \frac{dy}{dx} = 3\)

This simplifies to:

\(e^{x+y} + e^{x+y} \frac{dy}{dx} + \frac{dy}{dx} = 3\)

We can combine the terms involving \(\frac{dy}{dx}\):

\((e^{x+y} + 1) \frac{dy}{dx} = 3 - e^{x+y}\)

Now, solving for \(\frac{dy}{dx}\), we get:

\(\frac{dy}{dx} = \frac{3 - e^{x+y}}{e^{x+y} + 1}\)

Therefore, the correct answer is \(\frac{3-e^{x+y}}{e^{x+y}+1}\), which matches the correct option.

Thus, the correct answer is:

 

$\dfrac{3-e^{x+y}}{e^{x+y}+1}$