# Derivative question

• Mar 11th 2010, 11:59 AM
anderson
Derivative question
Hi everyone

Need help to verify this working.Thank you in advance for all help & support,really appreciate.

Find $\displaystyle \frac{dz}{dx}$&$\displaystyle \frac{dz}{dy}$ for
a) xy + yz - xz = 0
$\displaystyle y + y\frac{dz}{dx}-z-x\frac{dz}{dx}=0$
$\displaystyle \frac{dz}{dx}(y-x)=z-y$
$\displaystyle \frac{dz}{dx}=\frac{z-y}{y-x}$

$\displaystyle x+z+y\frac{dz}{dy}-x\frac{dz}{dy}=0$
$\displaystyle \frac{dz}{dy}(y-x)=-x-z$
$\displaystyle \frac{dz}{dy}=\frac{-x-z}{y-x}$

b) xyz-cos(x+y+z)
$\displaystyle F_x =yz+sin(x+y+z)$
$\displaystyle F_y=xz+sin(x+y+z)$
$\displaystyle F_z= xy+sin(x+y+z)$
$\displaystyle \frac{dz}{dx}=-\frac{F_x}{F_z}=-\frac{yz+sin(x+y+z)}{xy+sin(x+y+z)}$
$\displaystyle \frac{dz}{dy}=-\frac{F_y}{F_z}=-\frac{xz+sin(x+y+z)}{xy+sin(x+y+z)}$

c) $\displaystyle 4ln(4xyz) + sin(xz^2))$
$\displaystyle F_x=\frac{4}{x}+z^2cos(xz^2)$
$\displaystyle F_y=\frac{4}{y}$
$\displaystyle F_z=\frac{4}{z}+2xzcos(xz^2)$
*$\displaystyle \frac{dz}{dx}=-\frac{F_x}{F_z}=-\frac{\frac{4}{x}+z^2cos(xz^2)}{\frac{4}{z}+2xzcos (xz^2)}$
*$\displaystyle \frac{dz}{dy}=-\frac{F_y}{F_z}=-\frac{\frac{4}{y}}{\frac{4}{z}+2xzcos(xz^2)}$
*Not sure how to simplify both equations,just left in this form.
Thank you in advance for all your kind help & support, really hope someone can verify for me.
• Mar 11th 2010, 12:19 PM
pickslides
$\displaystyle xy + yz - xz = 0$

$\displaystyle xy = -yz + xz$

$\displaystyle xy = z(x-y )$

$\displaystyle \frac{xy}{x-y } = z$

$\displaystyle z=\frac{xy}{x-y }$

Therefore

$\displaystyle \frac{dz}{dx}= \frac{(x-y)y- xy}{(x-y)^2 }=\dots$

$\displaystyle \frac{dz}{dy}=\frac{(x-y)x- xy}{(x-y)^2 }=\dots$