Question

Differentiation of the equation y = x^2 + z^2 with respect to x , thanks .

Answer 1

2x

Answer 2

y=2x+2z dz/dx

Answer 3

@AravindG : Since he's calculating dy/dx, wouln't z just be a constant, therefore the derivative of it would be 0?

Answer 4

Well it was not mentioned that z is a constant ..If it is the answer will simply be 2x

Answer 5

@AravindG : Since he's calculating dy/dx, wouldn't it be assumed that Z is just a constant, since he's calculating partial derivative?

Answer 6

y = x^2 + z^2
dy/dx = 2x or 2x +2xdz/dx

Answer 7

2zdz/dx

Answer 8

@FutureMathProfessor : Your concern should be a question , however I have started calculus after a long time break ; can you post a question with relevant example.

Answer 9

@AravindG : Can you please give a proper explanation of PDE and ODE with regard to when to consider a variable and not a constant . Thanks.

Answer 10

This question has been answered, what more do you need?

Answer 11

@eSpeX -- Additional help , when to consider z a constant and not a variable .

Answer 12

If you are asked to take a partial derivative of an equation containing more than one variable, you consider the other variables constants during the derivation of the other variables.
If you are simply asked to take the derivative of an equation, it is proper to take the derivative of terms with other variables, with respect to x, and append that term with d<>/dx.

Answer 13

Which calc are you in, you will be learning about partials towards the end of II and into III.

Answer 14

An example ..

Answer 15

@eSpeX : The last one basically .

Answer 16

So you are in Calc III? I would expect that you will be covering this soon then.

Answer 17

Well to be fair , I was given to solve this equation dy/dx = sqrt( y- x) , and it arised from that .

Answer 18

An example using your equation. If you are asked to take the derivative of x^2 + z^2, you would write \[\frac{ dy }{ dx }=x^2 + z^2 \rightarrow 2x+2z\frac{dz}{dx}\] where as if you were asked to take the partial dz/dx and dy/dx of the same function, you would get \[\frac{dz}{dx}=x^2 + z^2\rightarrow 2z\]and\[\frac{dy}{dx}=x^2 + z^2\rightarrow 2x\]

Answer 19

The 'd' would be different, I did not see a partial diff symbol in the equation library.

Answer 20

I know about the 'd'

Answer 21

Now last question , a little bit theoretical if you are interested , I read about it a long time back ; what is the difference between these two 'd's.

Answer 22

As far as the origination I am uncertain.

Answer 23

No its terms of application , the partial differentiation and the normal differentiation , just to brush up my memory .

Answer 24

In terms of application they denote how you handle the function with multiple variables.

Answer 25

y = x^2 + z^2
z = sqrt(y -x^2)
why not the partial derivation
dz/dx =-2x/sqrt(y - x^2)

Answer 26

@KenLJW : We are not on the same page , may be .

Answer 27

slight error
dz/dx =-x/sqrt(y - x^2)
was responding to previous answer for partial derivative