Find dy dx by implicit differentiation. x2 x + y = y2 + 1 dy dx

To find $\frac{dy}{dx}$ by implicit differentiation, let’s start with the given equation:

$x^2(x + y) = y^2 + 1$

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

   The left side involves the product of two functions of $x$ and $y$: $x^2$ and $(x + y)$. We’ll apply the product rule here.

   $\frac{d}{dx} \left( x^2(x + y) \right) = \frac{d}{dx} (y^2 + 1)$

2. Apply the Product Rule on the left side:

   The product rule states that $\frac{d}{dx}(u \cdot v) = u'v + uv'$. Here, let $u = x^2$ and $v = x + y$.

   • Differentiate $u = x^2$: $u' = 2x$.
   • Differentiate $v = x + y$ with respect to $x$: $v' = 1 + \frac{dy}{dx}$.

   Now apply the product rule:

   $\frac{d}{dx} \left( x^2(x + y) \right) = (2x)(x + y) + (x^2)(1 + \frac{dy}{dx})$

3. Differentiate the right side:

   Now, differentiate $y^2 + 1$ with respect to $x$:

   $\frac{d}{dx}(y^2 + 1) = 2y \cdot \frac{dy}{dx}$

   Putting it all together:

   $(2x)(x + y) + x^2\left(1 + \frac{dy}{dx}\right) = 2y \cdot \frac{dy}{dx}$

4. Expand and simplify:

   Expanding the left side:

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

5. Isolate $\frac{dy}{dx}$:

   Bring terms involving $\frac{dy}{dx}$ to one side and terms without $\frac{dy}{dx}$ to the other side:

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

   Factor out $\frac{dy}{dx}$ on the left side:

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

6. Solve for $\frac{dy}{dx}$:

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

So, the solution is:

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

Would you like further details on any step?

Here are 5 related questions to deepen understanding:

1. How does implicit differentiation differ from explicit differentiation?
2. What is the product rule, and when do we use it in differentiation?
3. Can implicit differentiation be used for any implicit function?
4. How can we interpret $\frac{dy}{dx}$ graphically for an implicit function?
5. What are some practical applications of implicit differentiation?

Tip: When using implicit differentiation, keep track of each term’s dependency on $x$ and $y$ to avoid missing terms where $\frac{dy}{dx}$ applies.