- #1

- 457

- 0

the y is biglike the x its not small

and how do you differentiate this

xy^2

for the first one i used the product rule and thesecond one and i get a different answer then what i was suppost to get

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- Thread starter afcwestwarrior
- Start date

- #1

- 457

- 0

the y is biglike the x its not small

and how do you differentiate this

xy^2

for the first one i used the product rule and thesecond one and i get a different answer then what i was suppost to get

- #2

- 457

- 0

x^2 d/dx y dy/dx + y d/dx x^2

x^2 y' + y+ 2x

so how come im not getting 2x 2y

- #3

- 1,425

- 1

You need an equation, not an expression, i.e. [tex]x^{2}y = ?[/tex]. Also, are you familiar with implicit differentiation?

Last edited:

- #4

- 457

- 0

ill give u the equation x^3+x^2y+4y^2=6

- #5

- 457

- 0

- #6

- 1,425

- 1

[tex]x^3 + x^2y + 4y^2 = 6[/tex]

Okay, but do you know what implicit differentiation is? For example,

[tex]x + y = 1 [/itex]

[tex]\frac{dx}{dx} + \frac{dy}{dx} = \frac{d1}{dx} [/tex]

[tex]1 + \frac{dy}{dx} = 0 [/tex]

Also for a product, the product rule applies, e.g.

[tex]xy = 1[/tex]

[tex]x*\frac{dy}{dx} + \frac{dx}{dx}*y = 0 [/tex]

Okay, but do you know what implicit differentiation is? For example,

[tex]x + y = 1 [/itex]

[tex]\frac{dx}{dx} + \frac{dy}{dx} = \frac{d1}{dx} [/tex]

[tex]1 + \frac{dy}{dx} = 0 [/tex]

Also for a product, the product rule applies, e.g.

[tex]xy = 1[/tex]

[tex]x*\frac{dy}{dx} + \frac{dx}{dx}*y = 0 [/tex]

Last edited:

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