how is the differentiation of x=ky^2 equal to
$$1=2ky\frac{dy}{dx}$$

Question

how is the differentiation of x=ky^2 equal to 
$$1=2ky\frac{dy}{dx}$$

anonymous · Answer

implicite differentiation, in this case y(x), such that y is a function of x.

anonymous · Answer

But that's only one guess in this case, the mechanical way I remember for implicit differentiation is derivate as if you were deriving something in terms of x and then just multiply it by dy/dx, chain rule.

anonymous · Answer

I"m working on the separable equations section of my differential equations chapter.

anonymous · Answer

I will just add this, maybe it helps you, this is the chain rule for multivariable calculus. 

The above equation you can write like that 
$$z=f(x,y)=x-ky^2$$

So the multivariable chain rule says

$$ dz = f_x dx + f_ydy $$
$$ \frac{dz}{dx}= f_x+f_y\frac{dy}{dx}$$ 
This is far from a proof, but you can read some application out of it. Implicit differentiation doesn't selectively deal with partial derivatives though.

anonymous · Answer

The original question reads. Find the orthogonal trajectories of the family of curves x=ky^2, where k is an arbitrary constant. 
And the first thing they did was differentiate x=ky^2 to get 
$$1=2ky\frac{dy}{dx}$$

anonymous · Answer

the gradient would be orthogonal.

anonymous · Answer

I haven't learn anything about gradients yet. This is only chapter 9 of stewart's calculus

anonymous · Answer

Another relationship for orthogonal functions is

$$ m_n \cdot m_y = -1 $$ where $m_n$ is normal to $m_y$ but I don't see why they apply this sort of differentiation here.

anonymous · Answer

Oh I think I see what they've done. 
They rearranged the equation to a separable equation, did the integral. 
Then stated:
THe orthogonal trajectories are the family of ellipses given by the following equation. 
$$x^2+\frac{y^2}{2}=C$$

anonymous · Answer

for y=k/x
I get
ln|y|=ln|-x|+C
What do you think?

anonymous · Answer

is this the integrated form?

anonymous · Answer

yep. 
$$\int \frac1ydy=\int-\frac1x dx$$

anonymous · Answer

$$ \large \int \frac{1}{y}dy = - \int \frac{1}{x}dx $$

So you can distribute the minus sign and you don't need to carry it inside your logarithmn.

anonymous · Answer

ok
ln|y|=-ln|x|+C

anonymous · Answer

perfect.

anonymous · Answer

Thank you!