If sin y + e⁻ˣᶜᵒˢ y = e, then dy/dx at (1, π) is equal to

The correct option/answer is e.

To find the value of $\frac{dy}{dx}$ at the point $(1,\pi )$, we start with the given equation:

$siny+e^{-xcosy}=e$

We differentiate both sides of the equation with respect to $x$ using the chain rule and the product rule.
Let us differentiate each term step-by-step:
1. The derivative of $siny$ with respect to $x$ is:
$cosy·\frac{dy}{dx}$
2. The derivative of $e^{-xcosy}$ with respect to $x$ is:
$e^{-xcosy}·\frac{d}{dx}(-xcosy)$
Applying the product rule to $-xcosy$:
$\frac{d}{dx}(-xcosy)=-cosy-x(-siny)\frac{dy}{dx}=-cosy+xsiny\frac{dy}{dx}$
3. The derivative of the constant $e$ on the right-hand side is $0$.

Putting it all together, we get:
$cosy\frac{dy}{dx}+e^{-xcosy}\left[-cosy+xsiny\frac{dy}{dx}\right]=0$

Now, we substitute the coordinates of the given point $(x,y)=(1,\pi )$ into this equation:
At $x=1$ and $y=\pi$, we have:
• $cos\pi =-1$
• $sin\pi =0$
• $e^{-xcosy}=e^{-1·(-1)}=e^1=e$

Substituting these values into the differentiated equation:
$(-1)\frac{dy}{dx}+e\left[-(-1)+1·\left(0\right)\frac{dy}{dx}\right]=0$

Simplifying the expression:
$-\frac{dy}{dx}+e\left[1\right]=0$

$-\frac{dy}{dx}+e=0$

Rearranging to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx}=e$