Question

How would we find the derivative of this function?

Answer 1

\[x^{2}+y^{2} = 1 \]

Answer 2

supposed to find

\[\frac{ dy }{ dx }\]

Answer 3

sorry it's not a function because it's a circle

Answer 4

this is implicit differentiation

differentiate term by term

treat y as a function of x and use the chain rule

for example dy/dx for y^3 will be 3y^2 * dy/dx

Answer 5

find the derivative of y^2 in a similar way

Answer 6

the derivatives of x^ 2 and 1 are easy

Answer 7

sfter differentiating term by term you make dy/dx the subject of the equation

Answer 8

Okay, this is what I got so far

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

\[2x+\frac{ dy }{ dx }(y)^{2} = 0\]

Answer 9

but I don't understand how to do y^{2}

Answer 10

the seconf term is not correct - use the chain rule

y is a function of x

so we have dy/dx * 2y

Answer 11

suppose we ahd ( 2x- 3)^2

derivative is 2(2x - 3) * 2

- if 2x - 3 = y we proceed in similar fashion

Answer 12

I don't understand where the chain rule comes in

\[\frac{ dy }{ dx }*y^{2}\]

Answer 13

Yeah see I understand that you must take the derivative of both sides

Answer 14

@Michele_Laino help!

Answer 15

\[x^{2}+y^{2} = 1\]

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

\[2x+\frac{ dy }{ dx }(y)^{2} = 0\]

\[2x+\frac{ dy }{ dx }*2y = -2x\]

\[\frac{ dy }{ dx }2(y) = -2x\]

\[\frac{ dy }{ dx }= [\frac{ -x }{ y }]\]\]

Answer 16

But I don't understand where chain rule come in

Answer 17

When we have y = x^{2} and we take the derivative of y with respect to x, we're differentiating x

so

\[dy/dx = 2x = y'\]


but then we have y^{2} now we need to differentiate y but how can we do this the same way as we did before?

Answer 18

hmm the chain-rule comes in, to get dy/dx from the derivative of \(y^2\)
because when getting that derivative, you're really getting \(\bf \cfrac{d}{dx}(y)^2\implies \cfrac{d}{dx}[y^2]\cdot \cfrac{d}{dx}[y]\)

Answer 19

you don't chain-rule "x", because "x" is an independent variable
you do chain-rule "y", because "y" is NOT a variable per se, is a function, in "x" terms

Answer 20

so... when taking the derivative of the "y", you need to get the outer function's derivate and the inner, it just so happens that the inner function is 'y" by itself

Answer 21

okay so, we cant just take the derivative of y itself because it's a function right?

Answer 22

we can take x because in our problem it's a variable

Answer 23

\[y^{2} = \frac{ d }{ dx }[y]^{2}*\frac{ d }{ dx }y\]

Answer 24

that's correct

Answer 25

that we couldn't just take dy/dx for y^{2} because it's not a variable

when they say y is a function of x

it just means that

y = f(x) right?

Answer 26

yeap

Answer 27

ok okay I think i get it a little better now thanks

Answer 28

yw