Answer to Question #342100 in Calculus for kai

Question

Determine the second derivative of g(x)=sin(2x³-9x)

Answer 1

$$g'(x)=(\sin(2x^3-9x))'=(6x^2-9)\cos (2x^3-9x)$$

$$g''(x)=((6x^2-9)\cos (2x^3-9x))'$$

$$=12x\cos (2x^3-9x)-(6x^2-9)^2\sin (2x^3-9x)$$