Let f(x,y) = (−(4x+y))5. Then

[(∂2f)/(∂x∂y)] = ???
[(∂3f)/(∂x∂y∂x)] = ???
[(∂3f)/(∂x2∂y)] = ???

Question

Let f(x,y) = (−(4x+y))5. Then

[(∂2f)/(∂x∂y)]	= ???	
[(∂3f)/(∂x∂y∂x)]   = ???	
[(∂3f)/(∂x2∂y)]	= ???

myininaya · Answer

$$f=(-(4x+y))^5?$$

anonymous · Answer

yes

myininaya · Answer

$$f=-(4x+y)^5 $$
$$f_x=-5(4x+y)^{5-1}(4)=-20(4x+y)^4$$

myininaya · Answer

$$f_{xy}=-20 \cdot 4(4x+y)^{4-1}(1)=-80(4x+y)^3$$

myininaya · Answer

all you have to do is treat everything else like a constant except the one you are differentiating  with respect to

myininaya · Answer

$$f_{xyx}=-80(3)(4x+y)^{3-1}(4)$$

myininaya · Answer

you should try the last one

myininaya · Answer

i will be happy to check it

anonymous · Answer

why multiply that last one by 4 ?!

myininaya · Answer

chain rule

anonymous · Answer

so -900(4x+y)^2 ?

myininaya · Answer

partial derivative with respect to x of 4x+y is 4
parital derivative with respect to y of 4x+y is 1

anonymous · Answer

i know that ! but my question is how is that thing works !? [(∂^3f)/(∂x∂y∂x)]

myininaya · Answer

?

anonymous · Answer

i know its higher order but the signs for it are hella confusing

anonymous · Answer

so in the last one i do it considering x twice then with y one !?

myininaya · Answer

yep

myininaya · Answer

i already did f_x
now you have to find f_xx
then f_xxy

anonymous · Answer

ok cool thanks

myininaya · Answer

i like this notation more than the one above
it looks less messy i think lol