Partial derivatives: Chapter 14.3

if f(x,y,z)=e^(xy)*ln(z), find fx, fy, fz

fx=ye^(xy)lnz. fy=xe^(xy)lnz. fz=(e^(xy))/z

Find partial derivative of x and partial derivative of y if z is implicitly defined as a function of x and y by the equation: x³+y³+z³+6xyz=1

3x²+(3z²(dz/dx))+6yz+(6xy*(dz/dx)). dz/dx= (-x²-2yz)/(z²+2xy) partial derivative of y is dz/dx= (-y²-2xz)/(z²+2xy)

If z=f(x,y) then the notation for fx(x,y)=

Dxf

If z=f(x,y) then the notation for fy(x,y)=

Dyf

If f(x,y)=x³+x²y³-2y², find fy(2,1)

Holding x constant and differentiating with respect to y, fy(x,y)=3x²y²-4y. fy(2,1)=3*2²1²-4*1=8

If f(x,y)=x³+x²y³-2y², find fx(2,1)

Holding y constant and differentiating with respect to x, we get fx(x,y)=3x²+2xy³ and then fx(2,1)=3*2²+2*2*1³=16

A geometric interpretation of partial derivatives, we recall that the equation z=f(x,y) represents what graphically

It represents a surface S

What are the rules for finding partial derivatives of z=f(x,y)

To find fx, regard y as a constant and differentiate f(x,y) with respect to x. To find fy, regard x as a constant and differentiate f(x,y) with respect to y

If f(x,y)=4-x²-2y², find fx(1,1) and fy(1,1) and interpret these numbers as slopes

fx(x,y)=-2x. fy(x,y)=-4y. fx(1,1)=-2 fy(1,1)=-4

Find the second partial derivatives of f(x,y)=x³+x²y³-2y²

fx(x,y)=3x²+2xy³ fy(x,y)=3x²y²-4y then, fxx= (d/dx)(3x²+2xy³)= 6x+2y³ fxy= (d/dy)(3x²+2xy³)= 6xy² fyx= (d/dx)(3x²y²-4y)= 6xy² fyy= (d/dy)(3x²y²-4y)= 6x²y-4

If f is a function of two variables, its partial derivatives are the function fx(x,y) are defined

fx(x,y)=lim of ((f(x+h,y)-(f(x,y))/h as h goes to zero

fx(x,y,z) also known as the partial derivative with respect to x is defined as

fx(x,y,z)=lim of (f(x+h,y,z)-f(x,y,z))/h as h approaches zero

Calculate fxxyz if f(x,y,z)= sin(3x+yz)

fx=3cos(3x+yz) fxx=-9sin(3x+yz) fxxy=-9zcos(3x+ yz) fxxyz=-9cos(3x+ yz)+9yzsin(3x+yz)

Suppose f is defined on a disk D that contains the point (a,b). If the function fxy and fyz are both continuous on D, then

fxy(a,b)=fyx(a,b)

If f is a function of two variables, its partial derivatives are the function fy(x,y) are defined as

fy(x,y)= lim of (f(x,y+h)-f(x,y))/h

If g(x)=f(x,b) then what does fx(a,b)=

g'(a)

If g(x)=f(x,b) then what does the limit of fx(a,b)=

lim of ((f(a+h, b)-f(alb))/h) h goes to 0

If g(x)=f(x,b) then what does the limit of fy(a,b)=

lim of ((f(a,b+h) - f(a,b))/h) as h goes to zero

If f(x,y)=sin(x/(1+y)), calculate partial derivative of x

partial derivative of x= (cos(x/(1+y))*(d/dy(x/(1+y))= cos(x/(1+y))*(1/(1+y))

If f(x,y)=sin(x/(1+y)), calculate partial derivative of y

partial derivative of y= cos(x/(1+y)*(d/dy(x/(1+y))= -cos(x/(1+y))*(x/((1+y)²))