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

Step 1: Write the given relation.
The equation connecting the variables \(x\) and \(y\) is \[ e^{x+y}+y=3x \] Since \(y\) is a function of \(x\), we differentiate the equation implicitly with respect to \(x\).
Step 2: Differentiate each term.
Differentiate the exponential term \[ \frac{d}{dx}(e^{x+y}) \] Using the chain rule \[ \frac{d}{dx}(e^{x+y})=e^{x+y}\frac{d}{dx}(x+y) \] \[ = e^{x+y}\left(1+\frac{dy}{dx}\right) \] Differentiate \(y\) with respect to \(x\): \[ \frac{d}{dx}(y)=\frac{dy}{dx} \] Differentiate the right-hand side: \[ \frac{d}{dx}(3x)=3 \] Thus we obtain \[ e^{x+y}(1+\frac{dy}{dx})+\frac{dy}{dx}=3 \] Step 3: Expand the expression.
\[ e^{x+y}+e^{x+y}\frac{dy}{dx}+\frac{dy}{dx}=3 \] Group the derivative terms together: \[ e^{x+y}+(e^{x+y}+1)\frac{dy}{dx}=3 \] Step 4: Solve for $\dfrac{dy{dx}$.}
\[ (e^{x+y}+1)\frac{dy}{dx}=3-e^{x+y} \] Dividing both sides by \(e^{x+y}+1\): \[ \frac{dy}{dx}=\frac{3-e^{x+y}}{e^{x+y}+1} \] Step 5: Conclusion.
Thus the derivative of \(y\) with respect to \(x\) is \[ \frac{dy}{dx}=\frac{3-e^{x+y}}{e^{x+y}+1} \] Final Answer: $\boxed{\dfrac{3-e^{x+y}}{e^{x+y}+1}}$