Question

Partial differentiation
z=sqrt( x^3+y^3+3x^2y)

Answer 1

diff. w.r.t ?
to diff. w.r.t x, treat y as constant
to diff. w.r.t y, treat x as constant

Answer 2

The actual question sounds like this:
Show that,if \[z =\sqrt{x ^{3}+ y ^{3}+3x ^{2}y}\] then
\[x \frac{ dz }{ dx }+y \frac{ dz }{ dy }=\frac{ 3 }{ 2 }z\]

Answer 3

Hence deduce that
\[x ^{2}\frac{ d ^{2}z }{ dx ^{2} }+2xy \frac{ d ^{2}z }{ dxdy }+y ^{2}\frac{ d ^{2}z }{ dy ^{2} }\]
\[=\frac{ 3 }{ 4 }z\]
I manage to show the 1st order derivative bt stuck in 2nd part

Answer 4

which step ?

Answer 5

I stuck in showing the 2nd derivative

Answer 6

\[\frac{ \partial z }{ \partial x }=\frac{ 3x^2+6xy }{ 2 \sqrt{x^3+y^3+3x^2y} }\]\[\frac{ \partial z }{ \partial y }=\frac{ 3y^2+3x^2 }{ 2 \sqrt{x^3+y^3+3x^2y} }\]

Answer 7

Yeah,i gt this and able to show \[x \frac{ dz }{ dx }+y \frac{ dz }{ dy }=\frac{ 3 }{ 2 }z\]
Bt hw 2 deduce it?

Answer 8

I wouldn't bother deducing it!
The problem as stated is: show that x(dz/dx) + y(dz/dy) = 3z/2.
So calc the derivatives, multiply with x and y, and you get 3z/2. Done.

Now the other one: exactly the same, so first calculate the second order derivatives, multiply with x², 2xy and y² and "deduce" that this gives 3z/4.

As I understand it, your problem so far is to get the right 2nd order derivatives.
I'll give them a try myself now, so please be patient...

Answer 9

Partially differentiate \(x \frac{ dz }{ dx }+y \frac{ dz }{ dy }=\frac{ 3 }{ 2 }z\) w.r.t x
Then multiply both sides of that Equation with x.--->(1)
Partially differentiate \(x \frac{ dz }{ dx }+y \frac{ dz }{ dy }=\frac{ 3 }{ 2 }z\) w.r.t y
Then multiply both sides of that Equation with y----->(2).
From (1) and (2), you'll get your result.
@ZeHanz , taking 2nd derivative becomes very ugly, i've tried it and probably @jychay2 must also have tried it.

Answer 10

i get (1) as \(\large x\frac{ \partial^2 z }{ \partial x^2 }+ \frac{ \partial z }{ \partial x }+y \frac{ \partial^2 z }{ \partial x \partial y }=\frac{3}{2}\frac{ \partial z }{ \partial x }\)
u try n get (2)

Answer 11

after multiplying by x
\(\large x^2\frac{ \partial^2 z }{ \partial x^2 }+x \frac{ \partial z }{ \partial x }+xy \frac{ \partial^2 z }{ \partial x \partial y }=\frac{3}{2}x\frac{ \partial z }{ \partial x }--->(1)\)

Answer 12

Yeah, the 2nd derivatives lead me to nothing :(
But (implicitly) differentiating as you mentioned works. Here I calculate (2):

\[\frac{ d }{ dy }\left( x \frac{ dz }{ dx }+y \frac{ dz }{ dy } \right)=\frac{ 3 }{ 2 }\frac{ dz }{ dy }\]\[x \frac{ d^2z }{ dydx }+1\frac{ dz }{ dy }+y \frac{ d^2z }{ dy^2 }=\frac{ 3 }{ 2 }\frac{ dz }{ dy }\]Multiply both sides with y:\[xy \frac{ d^2z }{ dydx }+y \frac{ dz }{ dy }+y^2\frac{ d^2z }{ dy^2 }=\frac{ 3 }{ 2 }y\frac{ dz }{ dy }\]Now z behaves nicely, so \[\frac{ d^2z }{ dydx }=\frac{ d^2z }{ dxdy }\]so adding up (1) and (2) gives:\[x^2\frac{ \partial^2 z }{\partial x^2 }+x \frac{ \partial z }{ \partial x }+2xy \frac{ \partial^2z }{ \partial x \partial y }+y \frac{ \partial z }{ \partial y }+y^2\frac{ \partial^2 z }{ \partial y^2 }=\frac{ 3 }{ 2 }\left( x \frac{ \partial z }{ \partial x } +y \frac{ \partial z }{ \partial y }\right)\]Subtracting x(dz/dx) and y(dz/dy) and using the formula x(dx/dx+y(dz/dy)=3z/2 gives:\[x^2\frac{ \partial^2z }{ \partial x^2 }+2xy \frac{ \partial^2z }{ \partial x \partial y }+y^2\frac{ \partial^2 z }{ \partial y^2 }=\frac{ 3 }{ 4 }z\]

Phew! A lot of trouble anyway. @hartnn : thanks for the tip!

Answer 13

Wow!! Finally i get this!!! Thx a lot @hartnn and @ZeHanz :)

Answer 14

yw!