dh df df dg

1. The derivative rules for multivariable functions stated Theorem 10 on page 151 are analogous to derivative rules from single variable calculus. Example 1 (page 152) illustrates the quotient rule. Recall from single-variable calculus that if f(x) and g(t) are each a differentiable function from R1 to R1, and h = f(g(t)), then the chain rule tells us that dh df df dg — = — = — — or h  (t) = f (t) =f (g(t)) g  (t) . dt dt dx dt Now suppose f(x,y) is a differentiable function from R2 to R1 and c(t) = (x(t),y(t)) is a differentiable function (path) from R1 to R2. Consider the derivative of h(t) = f(c(t)) = foc(t) = f(x(t),y(t)) with respect to t, that is, dh df — = — . dt dt f change in f in x direction — x x y (x+x , y+y) (x , y) f change in f in y direction — y y x

2. f change in f in x direction — x x f dx  — — t x dt y (x+x , y+y) (x , y) f dy  — — t y dt f change in f in y direction — y y x dx change in x— t dt x = dy change in y— t dt y = f = total change in f= (change in f in x direction) + (change in f in y direction)  f dx — — t x dt f dy + — — t y dt fh — = — tt f dx — — x dt f dy + — — y dt Taking the limit of both sides as t 0, it can be shown that

3. fh — = — = tt f dx — — x dt f dy + — — y dt Taking the limit of both sides as t 0, it can be shown that dx ff— ——dt xy dy — dt dh f dxf dy — = — — + — — = dt x dty dt = Df(x,y) c(t) This chain rule can be extended in the natural way to a situation where f is a differentiable function from Rn to R1 and c(t) is a differentiable function (path) from R1 to Rn. For instance if h(t) = f(x(t),y(t),z(t)) , then dx — dt dy fff— —— —dt xyz dz — dt dh f dxf dyf dz — = — — + — — + — — = dt x dty dtz dt = Df(x,y,z) c(t)

4. f(u,v) = u2ev – uv3 x = cos t , y = sin t Find dh/dt where h(t) = f(x(t),y(t)) . dh f dxf dy — = — — + — — = dt x dty dt Using the chain rule, we have (2xey – y3)(–sin t) + (x2ey – 3xy2)(cos t) = –2(cos t)(sin t)esin t + sin4t+ (cos3t)esin t – 3(cos2t)(sin2t) . Note how the same result can be obtained by first expressing h(t) = f(x(t),y(t)) in terms of t and then differentiating. u = xeyz x = et , y = t , z = sin t Find du/dt . du u dxu dyu dz — = — — + — — + — — = dt x dty dtz dt Using the chain rule, we have

5. u = xeyz x = et , y = t , z = sin t Find du/dt . du u dxu dyu dz — = — — + — — + — — = dt x dty dtz dt Using the chain rule, we have eyzet + xzeyz(1) + xyeyz(cos t) = et(sin t)et + et(sin t)et(sin t)+ et t et(sin t)cos t = (1 + sin t +t cos t) et(1+sin t). Note how the same result can be obtained by first expressing u in terms of t and then differentiating.

6. Suppose f(u,v) is a differentiable function from R2 to R1, and g(x,y) = [u(x,y) , v(x,y)] is a differentiable function from R2 to R2. Then, the derivative matrix for h(x,y) = f(u(x,y),v(x,y)) = fg(x,y) is _ _ | hh | | — — | |_ xy _| D(fg) = _ _ | f uf vf uf v | = | — — + — —— — + — — | |_ u xv xu yv y _| uu —— xy vv — — xy ff = —— uv = DfDg

7. Suppose f(u,v) = [f1(u,v) , f2(u,v)] is a differentiable function from R2 to R2, and g(x,y) = [u(x,y) , v(x,y)] is a differentiable function from R2 to R2. Then, the derivative matrix for h(x,y) = [h1(x,y),h2(x,y)] = f(u(x,y),v(x,y)) = [f1g(x,y), f2g(x,y)] =fg(x,y) is _ _ | h1h1 | | —— | | xy | | | | h2h2 | | —— | |_ xy _| D(fg) = _ _ _ _ | f1f1 | | uu | | —— | | —— | = | uv | | xy | | | | | | | | | | f2f2 | | vv | | —— | | — — | |_ uv _| |_ xy _| = DfDg Look at the general chain rule stated in Theorem 11 on page 153.

8. f(u,v) = uv g(x,y) = (x2 – y2 , x2 + y2) Find the derivative matrix for h(x,y) = f(u(x,y),v(x,y)) = fg(x,y). 2 1 2 2 2 1 f : R Rg: R Rh: R R Using the chain rule, we have _ _ _ _ _ _ | hh | | ff | | uu | | — — | = | —— | | —— | |_ xy _| | uv | | xy | |_ _| | | | | | vv | | — — | |_ xy _| _ _ _ _ | | | | | v u | | 2x –2y | = | | | | |_ _| | | = | | | 2x 2y | | | |_ _|

9. _ _ | | | v (2x) + u (2x) v (–2y) + u (2y) | |_ _| _ _ | | = | (x2 + y2)(2x) + (x2 – y2)(2x) (x2 + y2)(–2y) + (x2 – y2)(2y)| |_ _| _ _ | | = | 4x3 –4y3 | . |_ _| Since f(u,v) = uv and g(x,y) = (x2 – y2 , x2 + y2) , it is easy to see that h(x,y) = (x2 – y2)(x2 + y2) = x4 – y4 , after which it is easy to see that _ _ _ _ | hh | | | | — — | = | 4x3 –4y3 | . |_ xy _| |_ _| Is it possible to define gf ? No

10. f(u,v,w) = u2 + v2 – w g(x,y,z) = (x2y , y2 , e–xz) Find the derivative matrix for h(x,y,z) = f(u(x,y,z),v(x,y,z),w(x,y,z)) = fg(x,y,z). 3 1 3 3 3 1 f : R Rg: R Rh: R R _ _ | uu u | | —— — | _ _ _ _ | xy z | | hhh | | fff | | | | — — — | = | — — — | | vv v | |_ xyz _| | uv w | | — — — | |_ _| | xy z | | | | ww w | | — — — | |_ xy z _| Using the chain rule, we have _ _ _ _ | | | | | 2u 2v –1 | | 2xyx2 0 | = | | | | |_ _| | 0 2y 0 | = | | | –ze–xz 0 –xe–xz | |_ _|

11. _ _ | | | 2u (2xy) + ze–xz 2u (x2) + 2v (2y) xe–xz | |_ _| _ _ | | = | 2x2y (2xy) + ze–xz 2x2y (x2) + 2y2 (2y) xe–xz| |_ _| _ _ | | = | 4x3y2 + ze–xz 2x4y+ 4y3xe–xz| |_ _| Since f(u,v,w) = u2 + v2 – w and g(x,y,z) = (x2y , y2 , e–xz) , it is easy to see that h(x,y,z) = x4y2 + y4 – e–xz , after which it is easy to see that _ _ _ _ | hhh | | | | — —— | = | 4x3y2 + ze–xz2x4y+ 4y3xe–xz | . |_ xyz _| |_ _| Is it possible to define gf ? No

12. f(u,v) = (2u– 8v , u2 , v4) g(x,y) = (x3 + 4 , 5x– y2) Find the derivative matrix for h(x,y) = f(u(x,y),v(x,y)) = fg(x,y) at the point (x0 , y0) = (2 , –3) . Df(u,v) = Dg(x,y) = 2 3 2 2 2 3 f : R Rg: R Rh: R R 2 –8 2u 0 0 4v3 3x2 0 5 – 2y (u0 , v0) = g(x0 , y0) = g(2 , –3) = (12 , 1) Dfg(2 , –3) = Df(12 , 1) Dg(2 , –3) =

13. –16–48 288 0 20 24 2 –8 24 0 0 4 12 0 = 5 6 Is it possible to define gf ? No