# Implicit Derivative approaches

Sorry for my excessive verboseness...

Here's the equation as given:

$$x = 10 + \sqrt{x^2 + y^2}$$

Here are my direct implicit steps without modifying original equation:

\eqalign{ \dfrac{\mathrm d}{\mathrm dx}\left(x\right)& = \dfrac{\mathrm d}{\mathrm dx}\left(10 + \sqrt{x^2 + y^2}\right)\\ \dfrac{\mathrm dx}{\mathrm dx}& = \dfrac{\mathrm d}{\mathrm dx}\left(10 + \sqrt{x^2 + y^2}\right)\\ 1 &= \dfrac{\mathrm d}{\mathrm dx}\left(10 + \sqrt{x^2 + y^2}\right)\\ 1 &= \dfrac{\mathrm d}{\mathrm dx}\left(10\right) + \dfrac{\mathrm d}{\mathrm dx}\left(\sqrt{x^2 + y^2}\right)\\ 1& = 0 + \dfrac{\mathrm d}{\mathrm dx}\left(\sqrt{x^2 + y^2}\right)\\ 1 &= \dfrac{\mathrm d}{\mathrm dx}\left(\sqrt{x^2 + y^2}\right)\\ 1 &= \dfrac{\mathrm d}{\mathrm dx}\left(\left(x^2 + y^2\right)^{1/2}\right)\\ 1 &= \dfrac12 \left(x^2 + y^2\right)^{-1/2} \cdot\dfrac{\mathrm d}{\mathrm dx}\left(x^2 + y^2\right)\\ 1 &= \dfrac 1{2 \sqrt{x^2 + y^2}}\cdot\dfrac{\mathrm d}{\mathrm dx}\left(x^2 + y^2\right)\\ 1 &= \dfrac 1{2 \sqrt{x^2 + y^2}}\cdot\left(\dfrac{\mathrm d}{\mathrm dx}\left(x^2\right) + \dfrac{\mathrm d}{\mathrm dx}\left(y^2\right)\right)\\ 1 &= \dfrac 1{2 \sqrt{x^2 + y^2}}\cdot\left(2x\cdot\dfrac{\mathrm d}{\mathrm dx}\left(x\right) + 2y\cdot\dfrac{\mathrm d}{\mathrm dx}\left(y\right)\right)\\ 1 &= \dfrac 1{2 \sqrt{x^2 + y^2}}\cdot\left(2x\cdot\dfrac{\mathrm dx}{\mathrm dx} + 2y\cdot\dfrac{\mathrm dy}{\mathrm dx}\right)\\ 1 &= \dfrac 1{2 \sqrt{x^2 + y^2}}\cdot\left(2x + 2y\cdot\dfrac{\mathrm dy}{\mathrm dx}\right)\\ 1 &= \dfrac {2x}{2 \sqrt{x^2 + y^2}} + \dfrac {2y}{2 \sqrt{x^2 + y^2}}\cdot\dfrac{\mathrm dy}{\mathrm dx}\\ 1 &=\dfrac x{\sqrt{x^2 + y^2}}+ \dfrac{y}{\sqrt{x^2 + y^2}}\cdot\dfrac{\mathrm dy}{\mathrm dx}\\ 1 -\dfrac x{\sqrt{x^2 + y^2}} &= \dfrac{y}{\sqrt{x^2 + y^2}}\cdot\dfrac{\mathrm dy}{\mathrm dx}\\ \sqrt{x^2 + y^2}\cdot\left(1 - \dfrac x{\sqrt{x^2 + y^2}}\right) &= y\cdot\dfrac{\mathrm dy}{\mathrm dx}\\ \sqrt{x^2 + y^2} - \dfrac{x\cdot\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2}} &= y\cdot\dfrac{\mathrm dy}{\mathrm dx}\\ \sqrt{x^2 + y^2} - x &= y\cdot\dfrac{\mathrm dy}{\mathrm dx}\\ \left[\dfrac{\sqrt{x^2 + y^2} - x}y\right] &= \dfrac{\mathrm dy}{\mathrm dx}}

This result matches http://symbolab.com

However, Wolfram gives:

$$\frac{dy}{dx}= \frac{-10}y$$

In an effort to get to Wolfram's result, I tried isolating y first:

\eqalign{ x &= 10 + \sqrt{x^2 + y^2}\\ 10 + \sqrt{x^2 + y^2} &= x\\ \sqrt{x^2 + y^2} &= x - 10\\ \sqrt{x^2 + y^2}^2 &= \left(x - 10\right)^2\\ \sqrt{x^2 + y^2}^2 &= \left(x - 10\right)\left(x - 10\right)\\ \sqrt{x^2 + y^2}^2 &= x^2 - 20x + 100\\ x^2 + y^2 &= x^2 - 20x + 100\\ y^2 &= x^2 - 20x + 100 - x^2\\ y^2 &= -20x + 100}

I'm guessing this next step could be problematic by not taking $\pm\sqrt n$ into account.

\eqalign{y &= \sqrt{-20x + 100}\\ &= \left(-20x + 100\right)^{1/2}}

Proceeding to implicit derivative processing:

\eqalign{ \dfrac{\mathrm d}{\mathrm dx}\left(y\right) &= \dfrac{\mathrm d}{\mathrm dx}\left(\left(-20x + 100\right)^{1/2}\right)\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac{\mathrm d}{\mathrm dx}\left(\left(-20x + 100\right)^{1/2}\right)\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac12 \cdot\left(-20x + 100\right)^{-1/2}\cdot\dfrac{\mathrm d}{\mathrm dx}\left(-20x + 100\right)\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac1{2\sqrt{-20x + 100}}\cdot\dfrac{\mathrm d}{\mathrm dx}\left(-20x + 100\right)\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac1{2\sqrt{-20x + 100}}\cdot\left(\dfrac{\mathrm d}{\mathrm dx}\left(-20x\right) + \dfrac{\mathrm d}{\mathrm dx}\left(100\right)\right)\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac1{2\sqrt{-20x + 100}}\cdot\left(20\dfrac{\mathrm d}{\mathrm dx}\left(x\right) + 0\right)\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac1{2\sqrt{-20x + 100}}\cdot\left(20\dfrac{\mathrm dx}{\mathrm dx}\right)\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac1{2\sqrt{-20x + 100}}\cdot20\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac{20}{2\sqrt{-20x + 100}}\\ \dfrac{\mathrm dy}{\mathrm dx} &= \dfrac{10}{\sqrt{-20x + 100}}}

I'm not sure if these next $\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2$ steps are allowed, but it did lead to a simpler result even though it never matched the Wolfram result:

\eqalign{ \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 &= \left(\dfrac{10}{\sqrt{-20x + 100}}\right)^2\\ \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 &= \dfrac{10^2}{\sqrt{-20x + 100}^2}\\ \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 &= \dfrac{100}{-20x + 100}\\ \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 &= \dfrac{100}{-20\left(x + 5\right)}\\ \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 &= -\dfrac5{x + 5}\\ \dfrac{\mathrm dy}{\mathrm dx} &= \sqrt{-\dfrac5{x + 5}}}

No matter what I try, I can't figure out at all how Wolfram got such a simple result.

So which is the true derivative, and proper approach?

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$$x=10+\left(x^{2}+y^{2}\right)^{\frac{1}{2}}$$

Differentiating on both sides leads to:

$$1=\frac{1}{2}\left(x^{2}+y^{2}\right)^{-\frac{1}{2}}\left(2x+2yy'\right)$$ Multiplying both sides with $\left(x^{2}+y^{2}\right)^{\frac{1}{2}}$ leads to:

$$\left(x^{2}+y^{2}\right)^{\frac{1}{2}}=x+yy'$$ The LHS can be recognized as $x-10$ and substituting this leads to: $$y'=-10y^{-1}$$

You did well in your first detailed deduction, but you 'forgot' to take the last step: write $-10$ for $\sqrt{x^{2}+y^{2}}-x$

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It's worse than forgot, never even knew to back substitute. First time I've ever come across this with derivatives. Thanks! – redthumb Jul 15 '14 at 16:05

In trying to match Wolfram by first isolating $y$, you obtained $$y^2 = -20x + 100 \implies \color{blue}{y = \sqrt{-20x + 100}}.$$

Then you also went on to obtain, in your last line following your expression for $y$ $$\;\dfrac{dy}{dx} = \dfrac{10}{\color{blue}{\sqrt{-20x + 100}}}.$$

Note that $$\frac{dy}{dx} = \frac{10}{\color{blue}{\sqrt{20x + 100}}} = \frac {10}{\color{blue}{y}}$$

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You're there with your first approach - since $\sqrt{x^2 + y^2}-x = -10$.

Easiest way to get this is from $y^2 = -20x + 100$ (check your working as you missed a minus sign):$$2y\frac{dy}{dx} = -20$$

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You can use the implicit function theorem instead, which is much faster. If we have an implicit function $f(x,y)=c$, then $$\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$ In this case, let $f(x,y)=\sqrt{x^2+y^2} -x=-10$. Then $$\frac{dy}{dx}=-\frac{\frac{x}{\sqrt{x^2+y^2}}-1}{\frac{y}{\sqrt{x^2+y^2}}}=\frac{\sqrt{x^2+y^2}-x}{y}=-\frac{10}{y}$$

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