Answer to Question #220719 in Algebra for Justice

Question

Determine the first order partial derivative of the following:
1. F(x, y) = cos(e^t) dt
2.f(x,y,z)=xy^2e^-xz

Answer 1

1.

Given that

$F(x,y)=\int_y ^xcos(e^t)dt$

Differentiating partially with respect to x we get

$\frac{\delta F}{\delta x}=\frac{\delta}{\delta x}\int_y^xcos(e^t)dt$

$=cos(e^x).\frac{\delta}{\delta x}(x)+cos(e^t).\frac{\delta}{\delta x}(y)................$

By leibritz theorem

$\frac{\delta F}{\delta x}=cos(e^x)$

now, again differenting partially with respect to y we get

$\frac{\delta F}{\delta y}=\frac{\delta}{\delta y}\int_y^xcos(e^t)dt$

$=cos(e^x).\frac{\delta}{\delta y}(x)-cos(e^y).\frac{\delta}{\delta y}(y)$

by Leibritz rule

$\frac{\delta F}{\delta y}=-cos(e^y)$

2.

Given $f(x,y,z)=xy^2e^{-xz}$

The first order derivative

$f(x,y,z)=xy^2e^-{xz}\\F_x=\frac{\delta}{\delta x}(xy^2e^{-xz})\\y^2\frac{\delta}{\delta x}(xe^{-xz})\\y^2[x\frac{\delta}{\delta x}e^{-xz}+e^{-xz}\frac{\delta}{\delta x} x]\\F_x=y^2[xe^{-xz}(-z)+e^{-xz}]$

$F_x(x,y,z)=e^{-xz}y^2(1-xz)$

$F_y=\frac{\delta}{\delta y}(xy^2e^{-xz})=F_y=2xye^{-xz}\space and \space F_z=\frac{\delta}{\delta z}(xy^2e^{-xz})$

$F_z=xy^2\frac{\delta}{\delta z}(e^{-xz})\\xy^2e^{-xz}\frac{\delta}{\delta z}(-xz)\\F_z=-x^2y^2e^{-xz}$