Question

Please solve this question ​

Answer 1

QUESTION:

y = x² sinx , find $y_2$

ANSWER:

d²y / dx² = 2 sin x + 4x cos x - x² sin x

GIVEN:

y = x² sinx

TO FIND:

d²y / dx²

FORMULAE:

dx/dx = 1

d/dx(uv) = uv' + vu'

d/dx (x²) = 2x

d/dx(sin x) = cos x

EXPLANATION:

dy /dx = d/dx (x²) sin x + d/dx(sin x) (x²)

dy /dx = 2x(sin x) dx/dx + x² (cos x ) dx/dx

dy /dx = 2x sin x + x² cos x

d²y / dx² = 2(dx/dx sin x + x d/dx(sin x)) + d/dx(x²) cos x + x² d/dx(cos x))

d²y / dx² = 2( sin x + x cos x) + 2x cos x + x²(- sin x)

d²y / dx² = 2 sin x + 2x cos x + 2x cos x - x² sin x

d²y / dx² = 2 sin x + 4x cos x - x² sin x

Hence $y_2$ = 2 sin x + 4x cos x - x² sin x