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$$ x^{a}y^{b} = c, $$

where a, b and c are constants. My attempts so far

$$ \frac{dy}{dx} = ax^{a - 1}by^{b - 1}$$

$$ \frac{d^2y}{dx^2} = (a^2 - a)x^{a-2}(b^2 - b)y^{b - 2} $$

I think that these first and second derivatives are correct, however my issue is, are these the derivatives in their simplest form?

Any hints or inputs are welcomed.

Ethan Bolker
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John Allen
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  • Do you mean the differential of $x^ay^b$ with respect to $x$, or do you mean you wish to find $\frac{dy}{dx}$? – mrnovice Mar 25 '17 at 18:15
  • The $ \frac{dy}{dx}$ was put in by me so I assume I have to find the differential with respect to x – John Allen Mar 25 '17 at 18:24

4 Answers4

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Hint: $$\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$$ Then differentiate the $\frac{d}{dx}$ similarly. Assuming you are not dealing with implicit differentiation otherwise you may need The Chain Rule as well as The Product Rule.

Nick Pavini
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$x^ay^b=c$

Using the product rule $$\frac{d}{dx}(u(x)v(x))=u'(x)v(x)+v'(x)u(x)$$

We get:

$$\frac{d}{dx}(x^ay^b)= ax^{a-1}y^b+by^{b-1}x^a\frac{dy}{dx} =0$$

$$\implies \frac{dy}{dx} = -\frac{ax^{a-1}y^b}{by^{b-1}x^a} , b\neq 0$$

We now use the quotient rule:

$$\frac{d}{dx}(\frac{u(x)}{v(x)}) = \frac{v(x)u'(x)-u(x)v'(x)}{v(x)^2}$$

Then $$\frac{d^2y}{dx^2} = -\frac{by^{b-1}x^a\cdot \frac{d}{dx}(ax^{a-1}y^b)-ax^{a-1}y^b\cdot \frac{d}{dx}(by^{b-1}x^a)}{(by^{b-1}x^a)^2}=0$$

Hopefully you can proceed from here.

mrnovice
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  • I can follow your explanation until the third equation – John Allen Mar 25 '17 at 18:39
  • If you let $u(x) = ax^{a-1}y^b$ and $v(x)=by^{b-1}x^a$, then plug it into the quotient rule formula, then the last line follows. – mrnovice Mar 25 '17 at 18:52
  • I am getting a long string of variables that dont seem to cancel. Should I see a lot of cancelling above the line? – John Allen Mar 25 '17 at 19:19
  • @JohnAllen There's no reason to suggest they should all cancel - however, David's answer will give a more elegant approach to doing it. – mrnovice Mar 25 '17 at 19:21
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Assuming this is implicit then, $$\frac{d}{dx}y^n=ny^{n-1}\frac{dy}{dx}$$

Use together with the product rule for correct solution

Brodie
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Take logs first

$$a\ln x+b\ln y=\ln c$$ $$\implies \frac ax+\frac by y'=0$$ $$\implies -\frac {a}{x^2}-\frac{b}{y^2}(y')^2+\frac by y''=0$$

Now you can substitute for $y'$ and get a very simple expression for $y''$

David Quinn
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