Question

Why is the derivative of \(y^2\) with respect to \(x\) equal to \(2y \cdot y'\) or \(2y \cdot \large{dy \over dx}\)?

I understand that the \(2y\) magic is power rule, but where does the \(y'\) come from?

Just to add, I do know that \({dy \over dx} = y'\) or \(y'(x)\) :)

Answer 1

this is the magic of chain rule !

Answer 2

Right, can you remind me what it really said?

Answer 3

derivative of y=x^2 is
\[\frac{dy}{dx}=2x*\frac{d}{dx}(x)=2x*1=2x\]
the same for y^2 produces
\[\Large 2y*\frac{d}{dx}(y)=2y \frac{dy}{dx}\]

Answer 4

implicit differentiation?

Answer 5

Yes, Omni.

Answer 6

I'm still trying to understand what @aliza_k said...

Answer 7

ok what dont you understand

Answer 8

I'm not really sure about the above.... wait a sec

Answer 9

that is not the chain rule..

Answer 10

power rule along with chain rule is
\[y=x^n\]
\[\frac{dy}{dx}=nx^{(n-1)}*\frac{d}{dx}(x)\]

Answer 11

u need to differentiate the base as well.

Answer 12

\[{d \over dx} f(g(x)) = f'(g(x))g'(x) \]

Answer 13

That one then?

Answer 14

Chain rule:
\[\frac{dy}{dt} =\frac{dy}{dx} *\frac{dx}{dt} \]
however, this is not required; only needed in related rates,

Answer 15

How to apply this rule to\[{d \over dx} y^2\]

Answer 16

Oh, got it... let me check.

Answer 17

for the one you to wrote; chain rule is needed.

Answer 18

\[\frac{d}{dx} y^2=\frac{d}{dy} \left(y^{2}\right) *\frac{dy}{dx} =2y*\frac{dy}{dx} \]

Answer 19

Sorry... and yes! I understood the application :)

Answer 20

\[{d \over dx}y^2 = 2y \times {dy \over dx}\]Yeah, so \(f\) is the square function and \(y(x)\) is the inner function :)

Answer 21

yes

Answer 22

\[y(x) = g\]

Answer 23

i wouldnt use functions hard to understand :)

Answer 24

... and then apply it to the rule I initially wrote

Answer 25

Thanks people!

Answer 26

thanks for your Question @ParthKohli

Useful for me too ;)