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

Mathematics
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implicite differentiation, in this case y(x), such that y is a function of x.
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.
I"m working on the separable equations section of my differential equations chapter.

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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.
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 \[1=2ky\frac{dy}{dx}\]
the gradient would be orthogonal.
I haven't learn anything about gradients yet. This is only chapter 9 of stewart's calculus
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.
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. \[x^2+\frac{y^2}{2}=C\]
for y=k/x I get ln|y|=ln|-x|+C What do you think?
is this the integrated form?
yep. \[\int \frac1ydy=\int-\frac1x dx\]
\[ \large \int \frac{1}{y}dy = - \int \frac{1}{x}dx \] So you can distribute the minus sign and you don't need to carry it inside your logarithmn.
ok ln|y|=-ln|x|+C
perfect.
Thank you!

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