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Prove Theorem when fis a function of three variables. That is, show that if x0,y0,z0is a point in the domain of f(x,y,z) at which the first-order partial derivatives off exist, and if u3 is a unit vector for which the directional derivative Dufx0,y0,z0 also exists, thenDufx0,y0,z0=fx0,y0,z0u.

Short Answer

Expert verified

Find out the gradient to prove the above relation.

Step by step solution

01

Given Information

Let f(x,y,z)be a function of three variables defined on a open set containing

x0,y0,z0and letu=,,the point containing a unit vector

The direction derivative in udirection is given by

Dufx0,y0,z0

Dufx0,y0,z0=limh0fx0+h,y0+h,z0+h-fx0,y0,z0h

limit also exists

The gradient of function is given by

fx0,y0,z0=fxx0,y0,z0i+fyx0,y0,z0j+fzx0,y0,z0k

As ,,,x0,y0andz0are constants

F(h)=fx0+h,y0+h,z0+his a single variable function

F(0)=fx0+0,y0+0,z0+0=fx0,y0,z0

02

Simplification

From above equation Dufx0,y0,z0=limh0F(h)-F(0)h=F'(0)

If x=x0+h,y=y0+handz=z0+h

By chain rule

F'(h)=fxxh+fyyh+fzzh

=fx(x,y,z)hx0+h+fy(x,y,z)hy0+h+fz(x,y,z)hz0+h

=fx(x,y,z)(0+)+fy(x,y,z)(0+)+fz(x,y,z)(0+)

=fx(x,y,z)+fy(x,y,z)+fz(x,y,z)

If h=0thenx=x0,y=y0andz=z0.

F'(0)=fxx0,y0,z0+fyx0,y0,z0+fzx0,y0,z0

Dufx0,y0,z0=fxx0,y0,z0+fyx0,y0,z0+fzx0,y0,z0

=fxx0,y0,z0,fyx0,y0,z0,fzx0,y0,z0,,

=fx0,y0,z0u

Hence proved

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