In Exercises 12.3 Questions 5 – 8, evaluate \(f_x(x, y)\) and \(f_y(x, y)\) at the indicated point.

Question 5:

\(f(x, y) = x^2y -x +2y +3\) at (1,2)

Answer :
To compute \(f_x(x, y)\) we keep y fixed.

\(f_x(x, y)= 2xy -1\)
Now substitute (1,2)
\(= 2*1*2 -1\)
=3

To compute \(f_y(x, y)\) we keep x fixed.
\(= x^2 +2\)
Now substitute (1,2)
\(= 1^2 +2 =3\)

Question 6:

\(f(x, y) = x^3 -3x+ y^2 -6y\) at (-1,3)

Answer :
To compute \(f_x(x, y)\) we keep y fixed. \(f_x(x, y)= 3x^2 -3\)
Now substitute (-1,3)
\(= 3*(-1)^2 -3 =0\)

To compute \(f_y(x, y)\) we keep x fixed. \(f_y(x, y)= 2y -6\)
Now substitute (-1,3)
\(2*3- 6 =0\)

Exercises 12.4

In Exercises 17 – 18, find the total differential dw

Question 17:

\(w = x^2yz^3\)

Answer : \(dw = f_x(x, y,z)dx + f_y(x, y,z)dy + f_z(x, y,z)dz\)

\(f_x(x, y,z)dx = 2xyz^3\)

\(f_y(x, y,z)dx = x^2z^3\)

\(f_z(x, y,z)dx = 3x^2yz^2\)

\(dw =2xyz^3 + x^2z^3 + 3x^2yz^2\)

Exercises 12.5

In Exercises 7 – 12, functions \(z = f(x, y), x = g(t)\) and \(y = h(t)\) are given.
(a) Use the Multivariable Chain Rule to compute \(\frac{dz}{dt}\).
(b) Evaluate \(\frac{dz}{dt}\) at the indicated t-value.

Question 7:

\(z = 3x + 4y, x = t^2, y = 2t; t = 1\)

Answer :

Multivariable Chain Rule

If \(z = f(x, y), x = g(t)\) and \(y = h(t)\), where f, g and h are differentiable functions.
Then \(z = f(x, y) = f(g(t), h(t))\) is a function of t, and

\(\frac{dz}{dt} =\frac{df}{dt} =f_x(x,y)\frac{dx}{dt} + f_y(x,y)\frac{dy}{dt}\)

Given , \(z = 3x + 4y, x = t^2, y = 2t;\)
\(z = 3x + 4y\)
\(\frac{dz}{dx} = 3\)
\(\frac{dz}{dy} = 4\)

\(\frac{dx}{dt} = 2t\)
\(\frac{dy}{dt} = 2\)

\(\frac{dz}{dt} = 3*2t + 4*2 =>6t +8\)

Evaluate \(\frac{dz}{dt}\) at t=1
\(=6*1 +8 =>14\)