VAIR¯AKU ARGUMENTU FUNKCIJU DIFERENCI¯ALR¯EK¸ INI

4. u = ey − 2x + 2, x = sin t, y = cos t, t0 = π
2 ;
5. u = x 2 e y , x = cos t, y = sin t, t0 = π;
6. u = ln(e x + e y ), x = t 2 , y = t 3 , t0 = 1;
7. u = x y , x = e t , y = ln t, t0 = 1;
8. u = e y−2x , x = sin t, y = t 3 , t0 = 0;
9. u = x 2 e −y , x = sin t, y = sin t 2 , t0 = π
2 ;
10. u = ln(e −x + e y ), x = t 2 , y = t 3 , t0 = −1;
11. u = e y−2x−1 , x = cos t, y = sin t, t0 = π
2 ;
12. u = arcsin x
y , x = sin t, y = cos t, t0 = π;
13. u = arccos 2x
y , x = sin t, y = cos t, t0 = π;
14. u = x2
y + 1 , x = 1 − 2t, y = arctg t, t0 = 0;
15. u = ln(e −x + e −2y ), x = t 2 , y = 1
3 t3 , t0 = 1.
IV Izskaitl¸ot dotās funkcijas f(x; y; z) parciālo atvasinājumu vērtības
punktā M0(x0; y0; z0) ar precizitāti līdz vienai simtdal¸ai:
1. f(x; y; z) =
z
x 2 + y 2 , M0(0; −1; 1);
2. f(x; y; z) = ln x + y
, M0(1; 2; 1);
2z
3. f(x; y; z) = (sin x) yz
π
, M0 ; 1; 2
6
4. f(x; y; z) = ln(x 3 + 2y 3 − z 3 ), M0(2; 1; 0);
5. f(x; y; z) =
x
y 2 + z 2 , M0(1; 0; 1);
6. f(x; y; z) = ln cos(x 2 y 2 + z),
M0 0; 0; π
;
4
7. f(x; y; z) = arctg(xy 2 + z),
2 x
8. f(x; y; z) = arcsin − z ,
y
9. f(x; y; z) =
M0(2; 1; 0);
M0(2; 5; 0);
√ z sin y
x , M0(2; 0; 4);
;
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