Question

Find f ''(x) when f(x)=sinx/ x^2. Thanks.

Answer 1

=cosx/x^2-2sinx/x^3

Answer 2

Thats wrong !

Answer 3

f'x = -2.1/x^3 *sinx + cosx*1/x^2

f"x = 6.1/x^4.sinx - 2*1/x^2 *cosx -sinx *1/x^2 - 3*1/x^3cosx

Answer 4

How do I simplify that? I dont get it!

Answer 5

ok

6 sinx 2cosx sinx 3cosx
------ - --------- - ---------- - --------
x^4 x^2 x^2 x^3

Answer 6

\[f'(x)=\frac{x^2\cos(x)-2x\sin(x)}{x^4}=\frac{x\cos(x)-2\sin(x)}{x^3}\]

Answer 7

\[f''(x)=-\frac{(x^2-6)\sin(x)+4x\cos(x)}{x^4}\] what a pain