Question

Find the partial derivative indicated. Assume the variables are restricted to a domain on which the function is defined.
z=(5x^2y^4−y^4) / (16xy−5)
Find the partial derivative indicated. Assume the variables are restricted to a domain on which the function is defined.
z=(5x^2y^4−y^4) / (16xy−5)
@Mathematics

Answer 1

d/dx((5 x^2 y^4-y^4)/(16 x y-5))
Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = 5 x^2 y^4-y^4 and v = 16 x y-5:
= ((16 x y-5) (d/dx(5 x^2 y^4-y^4))-(5 x^2 y^4-y^4) (d/dx(16 x y-5)))/(16 x y-5)^2
Differentiate the sum term by term and factor out constants:
= ((16 x y-5) (d/dx(5 x^2 y^4-y^4))-(5 x^2 y^4-y^4) (16 y (d/dx(x))+d/dx(-5)))/(16 x y-5)^2
= ((16 x y-5) (5 y^4 (d/dx(x^2))-d/dx(y^4))-(5 x^2 y^4-y^4) (16 y (d/dx(x))+d/dx(-5)))/(16 x y-5)^2
= ((16 x y-5) (5 y^4 (d/dx(x^2))-d/dx(y^4))-(5 x^2 y^4-y^4) (16 y (d/dx(x))+0))/(16 x y-5)^2
= ((16 x y-5) (5 (2 x) y^4-d/dx(y^4))-16 y (5 x^2 y^4-y^4) (d/dx(x)))/(16 x y-5)^2
= ((16 x y-5) (10 x y^4-d/dx(y^4))-16 1 y (5 x^2 y^4-y^4))/(16 x y-5)^2
= ((16 x y-5) (10 x y^4-0)-16 y (5 x^2 y^4-y^4))/(16 x y-5)^2
\[\frac{d}{dx}=\frac{2 y^4 (-25 x + 8 y + 40 x^2 y)}{(5 - 16 x y)^2}\]
That's your partial with respect to x.
with respect to y:
d/dy((5 x^2 y^4-y^4)/(16 x y-5))
Use the quotient rule, d/dy(u/v) = (v ( du)/( dy)-u ( dv)/( dy))/v^2, where u = 5 x^2 y^4-y^4 and v = 16 x y-5:
= ((16 x y-5) (d/dy(5 x^2 y^4-y^4))-(5 x^2 y^4-y^4) (d/dy(16 x y-5)))/(16 x y-5)^2
Differentiate the sum term by term and factor out constants:
= ((16 x y-5) (d/dy(5 x^2 y^4-y^4))-(5 x^2 y^4-y^4) (16 x (d/dy(y))+d/dy(-5)))/(16 x y-5)^2
= ((16 x y-5) (5 x^2 (d/dy(y^4))-d/dy(y^4))-(5 x^2 y^4-y^4) (16 x (d/dy(y))+d/dy(-5)))/(16 x y-5)^2
= ((16 x y-5) (5 x^2 (d/dy(y^4))-d/dy(y^4))-(5 x^2 y^4-y^4) (16 x (d/dy(y))+0))/(16 x y-5)^2
= ((16 x y-5) (5 x^2 (d/dy(y^4))-4 y^3)-16 x (5 x^2 y^4-y^4) (d/dy(y)))/(16 x y-5)^2
= ((16 x y-5) (5 x^2 (d/dy(y^4))-4 y^3)-16 1 x (5 x^2 y^4-y^4))/(16 x y-5)^2
= ((16 x y-5) (5 x^2 (4 y^3)-4 y^3)-16 x (5 x^2 y^4-y^4))/(16 x y-5)^2

\[\frac{d}{dy}=\frac{4 (5 x^2-1) y^3 (12 x y-5)}{(5 - 16 x y)^2}\]

Answer 2

I really appreciate the detailed work. Medal awarded!

Answer 3

It's hard to not be detailed with such things... just giving the answer to something that involved is rather useless...

Answer 4

Understandable, Im really struggling with this section haha

Answer 5

It take's a while to get used to the whole idea... I think it is what I struggled with the most throughout calc.