Question

Differentiate -7xy³

Answer 1

hmmm
Differentiate -7xy³
say p = -7 x y^3
then
dp = d/dx [ -7 x y^3] dx + d/dy [ -7 x y^3 ] dy
so
dp = [ -7 y^3 ] dx + [-21 x y^2 ] dy

Answer 2

To differentiate -7xy³, we can differentiate each term separately using the power rule.

The power rule states that if we have a term of the form cx^n, where c is a constant and n is a variable with a constant coefficient, we can differentiate it by multiplying the coefficient by the exponent and then subtracting 1 from the exponent.

Let's differentiate each term separately:

Differentiating -7:
The constant -7 does not have any variables, so its derivative is 0.

Differentiating x:
The variable x has an exponent of 1, so its derivative is 1.

Differentiating y³:
The variable y has an exponent of 3, so its derivative is 3y^2.

Putting it all together, the derivative of -7xy³ is:

0x + 1(-7)y³ + (-7x)3y²
Simplifying this expression, we get:

-7y³ - 21xy²

Therefore, the derivative of -7xy³ is -7y³ - 21xy².

Answer 3

Yes, that's correct! When differentiating with respect to multiple variables, we can use the partial derivative notation to indicate which variable we are differentiating with respect to. In this case, we have p = -7xy³ and we want to find dp.

Using the partial derivative notation, we have:
dp = (∂p/∂x) dx + (∂p/∂y) dy

Taking the partial derivatives, we find:
∂p/∂x = -7y³ (since ∂x/∂x = 1 and ∂y/∂x = 0)
∂p/∂y = -21xy² (since ∂x/∂y = 0 and ∂y/∂y = 1)

Substituting these results into the previous equation, we obtain:
dp = (-7y³) dx + (-21xy²) dy

So, dp = (-7y³) dx + (-21xy²) dy.