Animated Solution for Mathematics - Differentiation: If $e^y+xy=e$, the ordered pair $(\frac{dy}{dx},\frac{d^{2}y}{d{x}^2})$ at $x=0$ is equal to :

Welcome, my dear students. Today, we are not just solving a problem; we are peeling back the layers of an implicit function. When we see an equation like $e^y+xy=e$, it might look intimidating, but I want you to see it as a beautiful, hidden relationship between $x$ and $y$.

We are going to find the first and second derivatives at a specific point. If we stay disciplined, the algebra will unfold with elegance.

Before we even think about derivatives, we must ground ourselves. We are asked to evaluate our derivatives at $x=0$. However, the equation $e^y+xy=e$ requires us to know the corresponding $y$ value.

Let us substitute $x=0$ into our original equation:

The term $(0)y$ vanishes into thin air, leaving us with $e^y=e^1$. By the simple beauty of exponents, we immediately see that $y=1$. Our point of evaluation is $(0,1)$.

Now, let us perform the first differentiation. We differentiate the entire equation $e^y+xy=e$ with respect to $x$.

For the term $e^y$, we must respect the Chain Rule, yielding $e^y\cdot \frac{dy}{dx}$. For the term $xy$, we apply the Product Rule: $x\cdot \frac{dy}{dx}+y\cdot 1$. The derivative of the constant $e$ is zero.

Putting it together, we get:

Let us group the terms containing $\frac{dy}{dx}$:

Now, we substitute our point $(0,1)$ into this equation. With $x=0$ and $y=1$, we have:

This is where many students stumble, but you will not. We need $\frac{d^{2}y}{d{x}^2}$. We must differentiate our first derivative equation again:

We must apply the Product Rule to the first two terms. For $e^y\frac{dy}{dx}$, the derivative is:

For $x\frac{dy}{dx}$, the derivative is $x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}$. Combining these with the derivative of $y$, our full equation becomes:

Now, we substitute our known values: $x=0$, $y=1$, and $\frac{dy}{dx}=-\frac{1}{e}$.

Let us simplify this step-by-step:

Finally, we isolate $\frac{d^{2}y}{d{x}^2}$:

The final result for the second derivative is $\frac{1}{e^2}$. Remember, calculus is not about memorizing formulas; it is about the systematic application of rules.