Answer to Question #103879 in Calculus for Martin

Question

(1 point) Find d2ydx2 for
y=csc(x)
d2ydx2=

Answer 1

$y=csc(x)=1/sin(x)=sin^{-1}(x)$

using $(f(g(x)))'=f'(g(x))*g'(x)$

$y'=(-1)*sin^{-2}(x) *cos(x)$

using $f(g(x))'=f'(g(x))*g'(x)$ and $(fg)'=f'g+g'f$

$y''=(-1)* [ (-2)sin^{-3}(x) *cos(x)*cos(x) + sin^{-2}(x) *(-sin(x))]$

simplify:

$y''=2cot^2(x)csc(x)+csc(x)$

alternate form:

$y''=csc(x) (cot^2(x) + csc^2(x))$