Question

3 примера на вычисление производной сложной функции

Answer 1

Ответ:

Объяснение: 1) y=sin²(lg3x) + 1/2;

y"=(sin²(lg3x) + 1/2 )" = ( sin²(lg(3x))" + (1/2)" = (sin²(lg3x))" + 0=

=2×sin²⁻¹(lg3x) × (sin(lg3x))" = 2×sin(lg3x) × cos(lg3x) × (lg3x)" =

= sin(2lg3x) ×( (1×(3x)")/(3x×㏑10))=

=sin(2lg3x) × (3/(3×x㏑10))= (sin(2lg3x))/(x×㏑10).

2) y=sin²(xlg5x) + 1/∛x;

Находим производные поочередно каждой из составляющей суммы:

(sin²(xlg5x))" = 2×sin²⁻¹(xlg5x) × cos(xlg5x)×(xlg5x)" =

=2×sin(xlg5x) × cos(xlg5x) ×(x"×lg5x + x×(lg5x)" =

= sin(2xlg5x) × (lg5x + ((x×1)/(5x×㏑10)) =

= sin(2xlg5x) × (lg5x + (1/(5×㏑10)) =

= sin(2xlg5x) × lg5x + (sin(2xlg5x))/(5×㏑10);

(1/∛x)"= (1/(x^(1/3)))" =(x^(-1/3))" = -(1/3)×(x^(-1/3 -1)) =

= -(1/3)× (x^(-4/3)) = -1/(3×∛x⁴);

y"=(sin²(xlg5x) + (1/∛x))"=

=sin(2xlg5x)×lg5x +(sin(2xlg5x)/(5×㏑10)) - (1/(3×∛x⁴)).

3) y=√(cos(x/3)×(e^(tgx)));

y"= (√(cos(x/3)×(e^(tgx))))" =(1/2)×(cos(x/3) × e^(tgx))^(-1/2) ×

× (cos(x/3)× e^(tgx))" =

=(1/(2×√(cos(x/3)×e^(tgx))) × ((cos(x/3))"× e^(tgx) + cos(x/3)× ×(e^tgx)")=

=(1/(2×√(cos(x/3)×e^(tgx)) ×

×(-sin(x/3)×(x/3)"×e^(tgx) + cos(x/3)×(e^(tgx))×(tgx)")=

=  (1/((2×√((cos(x/3)×e^(tgx))) ×

×(-(1/2)×sin(x/3)×e^(tgx) + ((cos(x/3)×e^(tgx))/(2×cos²x×√(cos(x/3)×

×e^(tgx))=

=-(√(e^(tgx) × √(sin(x/3)) × √(sin(x/3))) / (4×√(cos(x/3) ) +

+ (√(e^(tgx))×√(cos(x/3))/(2×cos²x) =

= -(1/4)×√(sin(x/3)×tg(x/3)×e^(tgx)) +

+ (√((e^(tgx))×cos(x/3))/((2×cos²x)).