How to Complete Implicit Differentiation Problem

Question

The problem I am trying to find $\frac{dy}{dx}$ of: $$\sqrt{x+y}= x^4 + y^4$$

I have attempted to solve the problem via the following steps:

1. $x^{1/2} + y^{1/2} = x^4 + y ^4$
2. $\frac{d}{dx}x^{1/2}+\frac{d}{dx}y^{1/2} = \frac{d}{dx}x^4+\frac{d}{dx}y^4$
3. $\frac{1}{2}x^{-1/2}+\frac{1}{2}y^{-1/2}\frac{dy}{dx}=4x^3+4y^3\frac{dy}{dx}$
4. $\frac{1}{2}x^{-1/2}-4x^3=4y^3\frac{dy}{dx}-\frac{1}{2}y^{-1/2}\frac{dy}{dx}$
5. $\frac{1}{2}x^{-1/2}-4x^3=\frac{dy}{dx}(4y^3-\frac{1}{2}y^{-1/2})$
6. $\frac{\frac{1}{2}x^{-1/2}-4x^3}{4y^3-\frac{1}{2}y^{-1/2}}=\frac{dy}{dx}$

If someone could take a look at my steps and explain to me where I went wrong it would be greatly appreciated!

Answer 1

The error is in your first step, where you rewrite the equation $\sqrt{x+y} = x^4 + y^4$ as $x^{\frac{1}{2}}+y^{\frac{1}{2}} = x^4 + y^4$. As was pointed out in the comments $\sqrt{x+y} \not= x^{\frac{1}{2}}+y^{\frac{1}{2}}$ . For example $\sqrt{1+1} = \sqrt{2}$ which is not the same as $1^{\frac{1}{2}}+1^{\frac{1}{2}} = 2$.

You should note that the same is true of raising $(x+y)$ to other powers besides half, such as $(x+y)^2$, which is equal to $x^2 + 2xy + y^2$ not $x^2 + y^2$.

The way you should differentiate $\sqrt{x+y}$ is using the chain rule. First differentiate $\sqrt{x+y}$ with respect to $(x+y)$, which gives you $\frac{1}{2\sqrt{x+y}}$. Then, differentiate the internal expression $x+y$ with respect to $x$, which gives you $(1 + \frac{dy}{dx})$, then using the chain rule, we can determine that $\frac{d}{dx} \sqrt{x+y} = \frac{1}{2\sqrt{x+y}} \cdot (1 + \frac{dy}{dx})$ as was pointed out in the comments. I hope this helped.

Answer 2

Your first step in using $$x^{1/2}+y^{1/2}=x^4+y^4$$ is not correct. The problem is with $$x^{1/2}+y^{1/2}= \sqrt {x+y}$$

Make sure to fix that and the rest follows.