Use logarithmic differentiation:
y = 5(sin(x))x
ln(y) = ln(5(sin(x))x)
ln(y) = ln(5) + xln(sin(x))
differentiate both sides:
1/y•(dy/dx) = 0 + (1)(ln(sin(x)) + x•(1/sin(x))•cos(x)
1/y•(dy/dx) = ln(sin(x)) + xcot(x)
dy/dx = y[ln(sin(x)) + xcot(x)]
dy/dx = (5(sin(x))x)[ln(sin(x)) + xcot(x)]
plugging in x = 3 we get f '(3) = (5(sin(3))3)(ln(sin(3)) + 3cot(3))
f '(3) = 5•0.00281(ln(0.14112) - 21.04576)
f '(3) = 0.01405(-23.0039)
f '(3) = -0.323249
William W.
10/04/20
Precious W.
Its says incorrect answer. And I'm still a bit confused.10/04/20