Calculus: The Language of Change

Exercise Set 18.1
Partial and Total Derivatives

by David Theobald

Exercise 1 : Gradient Vectors

Given that the function z=f[x,y]=(x^2)(y^3) has partial derivatives, df/dx[x,y] = 2x(y^3) and df/dy[x,y]=3(x^2)(y^2):

part a

Give the gradient vector at a general point (x,y).

Solution of part a

The gradient vector is just made up of the partial derivatives of f with respect to both x and y.

∇f[X] = (({{df/dx[x, y]}, {df/dy[x, y]}})) = (({{2x(y^3)}, {3(x^2)(y^2)}}))

part b

Give the gradient at the specific point (x,y) = (1,1):

Solution of part b

To get the gradient at a specific point, all we must do is plug the values of the point into the equation discovered in part a.

If X = (1,1),  then:

∇f[X] = (({{df/dx[x, y]}, {df/dy[x, y]}})) = (({{2x(y^3)}, {3(x^2)(y^2)}})) = (({{2*1*(1^3)}, {3*(1^2)*(1^2)}})) = (({{2}, {3}}))

Exercise 2 : Total Differential

Given that the function z=f[x,y]=(x^2)(y^3) has partial derivatives, df/dx[x,y] = 2x(y^3) and df/dy[x,y]=3(x^2)(y^2):

part a

Give the total differential at a general point (x,y):

Solution of part a

The total differential is similar to implicit differentiation in that you multiply together the derivate with respect to x with dx and y with dy.  In this case, we can just take the dot product of the gradient found in exercise one with the perturbation vector X=[x,y] like so:

dz = <∇f[X]●dX>

This yields the following in this exercise:

dz = 2x(y^3)dx + 3(x^2)(y^2)dy

part b

Give the total differential at the specific point (1,1):

Solution of part b

To find the total differential at the given point, all we must do is plug the value of the point into the equation found in part a.

dz =  2*1*(1^3)*dx + 3*(1^2)*(1^2)*dy = 2dx + 3dy