How would I find the derivative of 2xy^2?

Question

Do you use the product rule? Can i break it up into 2x and y^2 to use the product rule?

Answer 1

It depends on which variable you are computing the derivative.

There is no such thing as "the derivative of a formula", but the "derivative of a function of one variable". If you take x and y as variables independent on each other, you can compute the derivative of the function that for a given x gives 2xy^2 for every y, or you can compute the derivative of the function that for a given y gives 2xy^2 for every x. The former is named "the partial derivative with respect to y", and its formula is "4xy", the latter is named "the partial derivative with respect to x", and its formula is "2y^2".

You can also take into account both partial derivatives; then you have a pair of functions, aka vector-valued function, named "gradient".

Edit: I assumed you had a function of two variables:

f: R^2 --> R, with f(x, y) = 2xy^2 for every x and for every y

and so you wanted to compute the following function:

D(f): R^2 --> R

This problem is not correctly specified.

You can have the partial derivative of f with respect to x:

Dx(f): R^2 --> R

or the partial derivative of f with respect to y:

Dy(f): R^2 --> R

or the gradient of f:

D(f): R^2 --> R^2

nidhi apparently assumed you had two functions of one variable:

g: R --> R, and represented "g(x)" with "y".

h: R^2 --> R, where h(x, y) = 2xy^2 for every x and for every y.

and their composition is taken:

hg: R --> R, where hg(x) = 2xg(x)^2

Then the derivative of this function has been computed:

D(hg): R --> R, where D(hg)(x) = 2g(x)(xD(g)(x)+g(x)) = 2y(xy'+y)

Answer 2

Yes, you would break it up into (2x)(y^2) and proceed to use the product rule.

Answer 3

2xy^2

2(xy^2)

use product rule on x and y^2

derivate with respect to x

2( xDy^2+y^2Dx)

2(x*2yy'+y^2*1)

2(2xyy'+y^2)

2y(xy'+y)