Question

Hi guys. I just started learning Derivatives, and solving Derivatives with one variable is easy.. but with 2 is making me confuse...
i got some sample questions here, and please try to explain how step by step. (i honestly don't know when to put in the dy/dx)

Instruction: Find dy/dx

x^3 + 2xy - y = 9

(y)/(xy - 1) = x

y = (5)^(3x)^2

Answer 1

Can you not just solve the equations for x and then take the derivative? For instance...

\[x^3 + 2xy-y=9\]
\[x^3+y(2x-1)=9\]
\[y=(9-x^3)/(2x-1)\]
then perform your derivative...

Answer 2

Sorry, I meant solve for y.

Answer 3

oh wait, forgot... the instruction was to find dy/dx...
well my teacher made this process. which confuses me alot

\[x^3y^2 + y =x\]

\[x^3(2y)(\frac{ dy }{ dx })+y^2(3x^2)+1(\frac{ dy }{ dx }) = 1\]

\[\frac{ dy }{ dx }(2x^3y+1)=-3x^2y^2\]

\[\frac{ dy }{ dx } = \frac{ -3x^2y^2 }{ 2x^3y+1 }\]

Answer 4

i just edited my question.... i must solve for dy/dx

Answer 5

So, the thing you have to remember is that y is a dependent variable (depended on x). Therefore, you treat all terms with y as having an x component. That's why why when you you did the first term of your example you solved it using the multiplication rule of derivatives.

Answer 6

Your teacher is using implicit differentiation, a very powerful technique.

Answer 7

\[x^3+2xy-y=9\]
\[3x^2+2y+2x(dy/dx)=0\]
\[dy/dx=(-2y-3x^2)/2x\]

Seem right?

Answer 8

Your goal here is not to solve y, but to find the derivative of it, dy/dx. I said its powerful because it let's you find an expression for the derivative dy/dx without having to solve y first.

Answer 9

so can you please explain briefly when is it the right time to place in the dy/dx

if \[x^3 = 3x^2\]

then what is y? or \[y^3\]

Answer 10

and if they are together like \[x^4y^3\]

Answer 11

You can always apply chain rule down to the variable you're differentiating with respect to.

Example:
\(\large\rm \frac{d}{dx}(x)^3=3(x)^2\frac{d}{dx}x\)
I'm applying chain rule, multiplying by the derivative of the inner function.
So we're left with: \(\large\rm =3x^2\frac{dx}{dx}\)

But notice that dx/dx doesn't really have any significance.
It's like asking, `how much does x change, as x changes?`

So we always suppress this dx/dx when differentiating with respect to the variable we're dealing with. It doesn't give us any extra information.

But I show you this just so you understand the pattern that the y's will follow.

Answer 12

So when we apply this idea to something like this:
\(\large\rm \frac{d}{dx}(y)^3=3(y)^2\frac{d}{dx}y\)
Now our chain actually gives us something that we care about.
\(\large\rm =3y^2\frac{dy}{dx}\)
so we do not suppress it.

And it all because we're differentiating with respect to some other variable.

Answer 13

When you're given something like \(\large\rm (x^4y^3)'\)
start by `setting up` your product rule,
\(\large\rm =(x^4)'y^3+x^4(y^3)'\)
And then deal with the derivatives one at a time.

And again, since my ' is d/dx, we're only ending up with these derivative functions when we differentiate y (because that's the one that doesn't match our derivative operator).
\(\large\rm =(4x^3)y^3+x^4(3y^2y')\)

Answer 14

You'll get product or chain rule cases as zepdrix has explained.
Independently though,

In general \[\frac{ d }{ dx } f(y) = \frac{ d }{ dy } f(y) \frac{ dy }{ dx }= f'(y) \frac{ dy }{ dx }\]

Answer 15

The equation I've written above illustrates why a dy/dx comes out as a factor when you differentiate an equation in y with respect to x.

Answer 16

Here are a couple of examples just to give you an idea. The latter is just an application of product rule.

\[\frac{ d }{ dx } (y^2) = \frac{ d }{ dy } (y^2) \frac{ dy }{ dx } = 2y \frac{ dy }{ dx }\]
\[\frac{ d }{ dx }(3y) = \frac{ d }{ dy }(3y) \frac{ dy }{ dx } = 3\frac{ dy }{ dx }\]
\[\frac{ d }{ dx } (xy) = x \frac{ d }{ dx }(y) + y \frac{ d }{ dx }(x) = x \frac{ dy }{ dx } + y\]
or alternatively
\[\frac{ d }{ dx }(xy) = \frac{ d }{ dy }(xy) \frac{ dy }{ dx } = (x \frac{ dy }{ dy } + y \frac{ dx }{ dy })\frac{ dy }{ dx }=x \frac{ dy }{ dx }+y\]