Question

Найти производную, как это решать? y=(arcsin4x)*e^x

Answer 1

d/dx(y) = d/dx(sin^(-1)(4 x) e^x)

0 = d/dx(e^x sin^(-1)(4 x))
d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), u = e^x v = sin^(-1)(4 x):
0 = sin^(-1)(4 x) d/dx(e^x)+e^x d/dx(sin^(-1)(4 x))
e^x is e^x:
0 = e^x (d/dx(sin^(-1)(4 x)))+e^x sin^(-1)(4 x)
, d/dx(sin^(-1)(4 x)) = ( dsin^(-1)(u))/( du) ( du)/( dx), u = 4 x ( d)/( du)(sin^(-1)(u)) = 1/sqrt(1-u^2):
0 = e^x sin^(-1)(4 x)+(d/dx(4 x))/sqrt(1-16 x^2) e^x

0 = e^x sin^(-1)(4 x)+(4 d/dx(x) e^x)/sqrt(1-16 x^2)

| 0 = e^x sin^(-1)(4 x)+(1 4 e^x)/sqrt(1-16 x^2)

Answer 2

$y'=(arcsin4x)' e^{x} + (e^{x})'(arcsin4x)= \frac{4e^{x}}{ \sqrt{1-16 x^{2} } } + (e^{x})(arcsin4x)$

$(e^{x})' = e^{x} \\ (arcsin4x)'=(4x)'(arcsin(4x))'= \frac{4}{ \sqrt{1- (4x)^{2} } }= \frac{4}{ \sqrt{1-16 x^{2} } }$