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The Derivative command can also be used to construct partial derivatives. Suppose f is a function of k variables, x1, x2, . . . , xk. nf Derivative[n1, n2,..., nk][f] gives the partial derivative n n where x1 x2... nk xk 1 2 n1 + n2 + ... + nk = n. It returns a pure function (see the appendix) that may then be evaluated at . [x1, x2,..., xk]
EXAMPLE 3 (Continuation of Example 2)
f[x_, y_] = x5 Y7; g = Derivative[3, 4][f] 50 400 #12 #23 & g[x, y] 50 400 x2 y3
Although the command D can be used to evaluate partial derivatives at a given point, Derivative is perhaps a bit more convenient.
EXAMPLE 4 Let f(x, y) = x3 sin y. Evaluate fxy at the point (2, ).
f[x_, y_] = x3 Sin[y]; D[f[x, y], x, y] /. {x 2, y o} 12 Derivative[1, 1][f][2, o] 12
SOLVED PROBLEMS
10.1 Compute the first- and second-order partial derivatives of f(x, y) = xexy.
SOLUTION
f[x_, y_] = x Exp[x y]; D[f[x, y], x] x y + x y x y D[f[x, y], y] x y x2 D[f[x, y], {x, 2}] 2 x y y + x y x y2 D[f[x, y], {y, 2}] x y x3 D[f[x, y], x, y] 2 x y x + x y x2 y
10.2 The partial derivatives of f(x, y) are defined by the following limits:
fx ( x , y) = lim
f ( x + h , y) f ( x , y) h f ( x , y + h ) f ( x , y) h
f y ( x , y) = lim
h 0
Multivariate Calculus
Compute the derivatives of f ( x , y) = ln( x 2 + y3 ) using the definition and verify using the Mathematica D command.
SOLUTION
f[x_, y_] = Log[x2 + y3]; Limit 2x x2 + y 3 f[x + h, y] f[x, y] , h 0 h
D[f[x, y], x] 2x x2 + y 3 Limit 2 3y x2 + y 3 f[x, y + h] f[x, y] , h 0 h
D[f[x, y], y] 3 y2 x2 + y 3
10.3 Let z = ex y. Compute
SOLUTION
3 z . x y
z = Exp[x y]; D[z, {x, 2}, y] 2 x y y + x y x y2
2 2 2 10.4 Verify that u = e a k t sin kx is a solution of the heat equation: u = a 2 u . t x 2
{x, 2}, y z
SOLUTION
u[x_, t_] = Exp[ a2 k2 t] Sin[k x]; lhs = D[u[x, t], t] a2 a2 k2 t k2 Sin [k x] rhs = a2 D[u[x, t], {x, 2}] a2 a2 k2 t k2 Sin [k x] lhs rhs True
10.5 A function of three variables, f(x, y, z), is said to be harmonic if it satisfies Laplace s equation: 1 2 f 2 f 2 f . Compute fxx, fyy, and fzz and show that f is Let f ( x , y, z ) = 2 2 + 2 + 2 = 0. x + y2 + z 2 x y z harmonic.
SOLUTION
f[x_, y_,z_]=
1 ; x2 + y 2 + z2
{x,2} f[x, y,z] / /Together
Multivariate Calculus
2x2 y 2 z2 5/2 (x2 + y 2 + z2) {y,2}f[x, y,z] / /Together o x2 + 2y 2 z2 5/2 (x2 + y 2 + z2) {z,2}f[x, y,z] / /Together x2 y 2 + 2z2 5/2 (x2 + y 2 + z2) %%% + %% + % / /Together 0
10.6 The plane tangent to the surface defined by z = f(x, y) at the point (x0, y0, z0) is z = z0 + f x ( x 0 , y0 )( x x 0 ) + f y ( x 0 , y0 )( y y0 ) Determine the equation of the plane tangent to the paraboloid z = 10 x2 2y2 at the point where x = 1 and y = 2. Sketch the paraboloid and its tangent plane.
SOLUTION
f[x_, y_] = 10 x2 2 y2; z = f[1, 2] + Derivative[1, 0][f][1, 2](x 1) + Derivative[0, 1][f][1, 2](y 2)//Expand 19 2x 8y
The tangent plane has equation z = 19 2 x 8 y.
g1 = Plot3D[f[x, y], {x, 5, 5}, {y, 5, 5}]; g2 = Plot3D[z, {x, 5, 5}, {y, 5, 5}] Show[g1, g2, PlotRange All, ViewPoint {2.330, 2.223, 1.040}]
50 0 50 5 0 5 5 0 5
10.7 The plane tangent to the surface f(x, y, z) = 0 at the point (x0, y0, z0) is f x ( x 0 , y0 , z0 )( x x 0 ) + f y ( x 0 , y0 , z0 )( y y0 ) + fz ( x 0 , y0 , z0 )( z z0 ) = 0 Sketch the sphere x 2 + y 2 + z 2 = 14 and its tangent plane at the point (1, 2, 3).
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