Differentiating $ x^{a}y^{b} = c $, in its simplest form. $$ x^{a}y^{b} = c,  $$
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.
 A: 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.
A: $x^ay^b=c$
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.
A: 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
A: 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 $y''$
