$$ \left\{\begin{array}{l} ext { Implicit disferemation } \ ext { b) } e^{x^{2} y}=x\end{array} ight. $$

Question

$$ \left\{\begin{array}{l}	ext { Implicit disferemation } \ 	ext { b) } e^{x^{2} y}=x\end{array}ight. $$

UpStudy AI · Accepted Answer

To perform implicit differentiation on the equation $$ e^{x^{2} y} = x $$, we will differentiate both sides with respect to $$ x $$. Here are the steps:

1. **Differentiate both sides**: We will apply the chain rule and product rule where necessary.

2. **Apply the chain rule**: For the left side, we have $$ e^{x^2 y} $$. The derivative of $$ e^u $$ is $$ e^u \cdot \frac{du}{dx} $$, where $$ u = x^2 y $$.

3. **Differentiate $$ u = x^2 y $$**: We will use the product rule since $$ u $$ is a product of $$ x^2 $$ and $$ y $$.

4. **Set the derivatives equal**: After differentiating both sides, we will solve for $$ \frac{dy}{dx} $$.

Let's go through these steps in detail.

### Step 1: Differentiate both sides

Differentiating the left side:
$$
\frac{d}{dx}(e^{x^2 y}) = e^{x^2 y} \cdot \frac{d}{dx}(x^2 y)
$$

Differentiating the right side:
$$
\frac{d}{dx}(x) = 1
$$

### Step 2: Differentiate $$ u = x^2 y $$

Using the product rule:
$$
\frac{d}{dx}(x^2 y) = x^2 \frac{dy}{dx} + 2xy
$$

### Step 3: Combine the results

Now we can substitute back into our differentiation:
$$
e^{x^2 y} \cdot (x^2 \frac{dy}{dx} + 2xy) = 1
$$

### Step 4: Solve for $$ \frac{dy}{dx} $$

Now we will isolate $$ \frac{dy}{dx} $$:
$$
e^{x^2 y} \cdot x^2 \frac{dy}{dx} + 2xy e^{x^2 y} = 1
$$
$$
e^{x^2 y} \cdot x^2 \frac{dy}{dx} = 1 - 2xy e^{x^2 y}
$$
$$
\frac{dy}{dx} = \frac{1 - 2xy e^{x^2 y}}{e^{x^2 y} \cdot x^2}
$$

Thus, the result of the implicit differentiation is:
$$
\frac{dy}{dx} = \frac{1 - 2xy e^{x^2 y}}{e^{x^2 y} \cdot x^2}
$$
$$ \frac{dy}{dx} = \frac{1 - 2xy e^{x^2 y}}{e^{x^2 y} \cdot x^2} $$

\( \left\{\begin{array}{l}\text { Implicit disferemation } \\ \text { b) } e^{x^{2} y}=x\end{array}\right. \)

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