calc_tutorial_14_6_014

 
Problem Statement

{3,5,7}
{3,4,5,6}
{2,4}
4*3-8
4*4
4*pow(5,3)+4*5+16

Use the Chain Rule to evaluate the partial derivative at the point specified.

(∂g / ∂s) at s = 5, where g(x, y) = x2 − y2, x = s2 + 3, y = 4 − 2s.

 
Step 1

Question Sequence

Question 1

Recall the General Version of the Chain Rule for functions of n variables. Let f(x1, , xn) be a differentiable function of n variables. Suppose that each of the variables x1, ..., xn is a differentiable function of m independent variables t1, ..., tm. Then we have the following for k = 1, ..., m.

(∂f / ∂tk) = (∂f / ∂x1) · (∂x1 / ∂tk) + (∂f / ∂x2) · (∂x2 / ∂tk) + ... + (∂f / ∂xn) · (∂xn / ∂tk)

We wish to find (∂g / ∂s) at s = 5 where g(x, y) is a differentiable function of two variables, x and y, and both x and y are differentiable functions of one variable, s.

Therefore we will apply the chain rule in the following way to calculate (∂g / ∂s).

A.
B.
C.
D.
E.

Note we have used the notation (dx / ds) and (dy / ds) instead of (∂x / ∂s) and (∂y / ∂s) since x and y are functions of only one variable, s. Otherwise, we would keep the notation for partial derivatives.

Find (∂g / ∂x) and (∂g / ∂y) in terms of x and y given that g(x, y) = x2 − y2.

(∂g / ∂x) =

A.
B.
C.
D.

(∂g / ∂y) =

A.
B.
C.
D.
Correct.
Incorrect.

 
Step 2

Question Sequence

Question 2

Find (dx / ds) and (dy / ds) given that x = s2 + 3 and y = 4 − 2s.

(dx / ds) =

A.
B.
C.
D.

(dy / ds) =

A.
B.
C.
D.
Correct.
Incorrect.

 
Step 3

Question Sequence

Question 3

Find (∂g / ∂s) in terms of only s by using the information above.

(∂g / ∂s) = (∂g / ∂x) · (dx / ds) + (∂g / ∂y) · (dy / ds)

= (2x)(2s) + (−2y)(−2)

= (2(s2 + 3))(2s) + (-2(4 − 2s))(-2)

= s3 + s +

Correct.
Incorrect.

 
Step 4

Question Sequence

Question 4

Find (∂g / ∂s) at s = 5 by substituting s = 5 into the expression for (∂g / ∂s).

Correct.
Incorrect.