Question

What is the derivative of x^2-2xy+y^3 ?

Answer 1

Which variable are you taking the derivative with respect to?

Answer 2

I'm not sure. We're learning Implicit Cost and the first step is to take the derivative, here is the full equation--> x^(2)-2xy+y^(3)=c

Answer 3

My teacher said to use the product rule and the first step was this 2x-[2y+2xy']+3y^(2)y' but how did he get that?

Answer 4

Lets take the implicit derivative with respect to x, and see what we get

\[f(x)=x^2-2xy+y^3=c\]
\[f'(x)=\frac{\mathrm d}{\mathrm dx}(x^2)-2\frac{\mathrm d}{\mathrm dx}(xy)+\frac{\mathrm d}{\mathrm dx}(y^3)=\frac{\mathrm d}{\mathrm dx}(c)%f'(x)=2x-2y-2xy'+3y^2y'\]

Answer 5

What do you get for this derivative?\[\frac{\mathrm d}{\mathrm dx}(x^2)=\]

Answer 6

And for these derivatives?

\[\frac{\mathrm d}{\mathrm dx}(xy)\]
\[\frac{\mathrm d}{\mathrm dx}(y^3)=\]
\[\frac{\mathrm d}{\mathrm dx}(c)=\]

Answer 7

for x^2 the derivative is 2x
derivative for xy is y' ?
derivative for y^3 is 3y^2
and for c my teacher just plugged in 0. Am I right in the derivatives?

Answer 8

you have the first term right

(x^2)' = 2x

Answer 9

for the second one you have to use the chain rule

\[\frac{\mathrm d}{\mathrm dx}(xy)=\frac{\mathrm d}{\mathrm dx}(x)y+x\frac{\mathrm d}{\mathrm dx}(y)=\]

Answer 10

\[=\frac{\mathrm dx}{\mathrm dx}y+x\frac{\mathrm dy}{\mathrm dx}=\]

Answer 11

can you simplify this, using the prime (') notation ?

Answer 12

(and remember that dx/dx = 1)

Answer 13

d/dx(xy)
x.dy/dx+y

Answer 14

and derivative of thrid term is 3y^2.dy/dx

Answer 15

if take derivative of y
\[\frac{ d }{ dx }(y)=\frac{ dy }{ dx }\]

Answer 16

\[f(x)=\frac{ d }{ dx }(x ^{2})-2\frac{ d }{ dx }(xy)+\frac{ d }{ dx }(y ^{3})\]

Answer 17

Actually I think I'm understanding it a bit more. I just have to stare at it for a while to get it into my head. The derivatives was the only thing that was confusing me. Thank you both for your help :)