# Thread: Another partial derivative problem...

1. ## Another partial derivative problem...

If f(x,y,z) = (x^3 + y^2 + z)^4 find each of the following:

a) fx(x,y,z)
b) fy(0,1,1)
c) fzz(x,y,z)

For a) I got 4(x^3 + y^2 + z)(3x^2)

b) 4(x^3 + y^2 + z)(2y) which then comes out to 8 once you plug in everything

Now c) is where I am lost. I have no clue how to differentiate fzz

2. Differentiate twice with respect to z:

1) Regard x and y as constants and use the chain rule:

u = z + x^3 + y^2
f(u)= u^4
f'(u)*u'= 4(z + x^3 + y^2)^3 * 1

f_z = 4(z + x^3 + y^2)^3

2) Chain rule again:

u = z + x^3 + y^2
f(u)= 4(u^4)
f'(u)*u'= 3*4(z + x^3 + y^2)^2 * 1

f_zz = 12(z + x^3 + y^2)^2

3. Hello, pakman!

If $f(x,y,z) \:= \:(x^3 + y^2 + z)^4$, find each of the following:

. . $(a)\; f_x(x,y,z)\qquad(b)\;f_y(0,1,1)\qquad(c)\; f_{zz}(x,y,z)$

Now (c) is where I am lost. .I have no clue how to differentiate $f_{zz }$
Just as $f_z$ means "differentiate with respect to $z$",
. . $f_{zz}$ means "do it twice".

We have: . $f \;=\;(x^3+y^2+z)^4$

Then: . $f_z\;=\;4(x^3+y^2+z)^3\cdot1 \;=\;4(x^3+y^2+z)^3$

Finally: . $f_{zz}\;=\;3\cdot4(x^3+y^2+z)^2\cdot1 \;=\;12(x^3+y^2+z)^2$