We are tasked with finding $\frac{dy}{dx}$ by implicit differentiation for the equation:

$x^5 - xy + y^2 = 8$

Step 1: Differentiate both sides with respect to $x$

When differentiating, apply the chain rule where necessary, especially since $y$ is a function of $x$.

1. Differentiate $x^5$ with respect to $x$: $\frac{d}{dx}(x^5) = 5x^4$

2. Differentiate $-xy$ with respect to $x$ (using the product rule): $\frac{d}{dx}(-xy) = -\left(\frac{d}{dx}(x) \cdot y + x \cdot \frac{dy}{dx}\right) = -\left(y + x\frac{dy}{dx}\right)$

3. Differentiate $y^2$ with respect to $x$ (using the chain rule): $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$

4. Differentiate the constant 8 with respect to $x$: $\frac{d}{dx}(8) = 0$

Step 2: Combine the results

Now, substitute all the differentiated terms back into the equation:

$5x^4 - (y + x\frac{dy}{dx}) + 2y \cdot \frac{dy}{dx} = 0$

Step 3: Group terms involving $\frac{dy}{dx}$

To isolate $\frac{dy}{dx}$, first rearrange the terms:

$5x^4 - y - x\frac{dy}{dx} + 2y \frac{dy}{dx} = 0$

Move the terms without $\frac{dy}{dx}$ to the other side:

$-x\frac{dy}{dx} + 2y\frac{dy}{dx} = y - 5x^4$

Step 4: Factor out $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the left-hand side:

$\frac{dy}{dx}(-x + 2y) = y - 5x^4$

Step 5: Solve for $\frac{dy}{dx}$

Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{y - 5x^4}{2y - x}$

Thus, the derivative is:

$\frac{dy}{dx} = \frac{y - 5x^4}{2y - x}$

Would you like further clarification or any additional details?

Related Questions:

1. How is implicit differentiation used in higher dimensions?
2. Can you explain the product rule in detail?
3. How do we apply the chain rule in implicit differentiation?
4. What happens if $y$ is a function of more than one variable?
5. How do you find the second derivative using implicit differentiation?

Tip:

When solving implicit differentiation problems, always remember to use the chain rule for terms involving $y$.