Based on the derivative on inverse trigonometric functions we have:
#color(blue)(d/dx(cos^-1u(x))=-(d/dx(u(x)))/(sqrt(1-u(x)^2))#
So, let us find #d/dx(u(x))#
Here ,#u(x)# is a composite of two functions so we should apply chain rule to compute its derivative.
Let
#g(x)=-2x^3-3# and
#f(x)=x^3#
We have #u(x)=f(g(x))#
The chain rule says:
#color(red)(d/dx(u(x))=color(green)(f'(g(x)))*color(brown)(g'(x))#
Let us find #color(green)(f'(g(x))#
#f'(x)=3x^2# then,
#f'(g(x))=3g(x)^2#
#color(green)(f'(g(x))=3(-2x^3-3)^2#
Let us find #color(brown)(g'(x))#
#color(brown)(g'(x)=-6x^2)#
#color(red)((du(x))/dx)=color(green)(f'(g(x)))*color(brown)(g'(x))#
#color(red)((du(x))/dx)=color(green)(3(-2x^3-3)^2)*(color(brown)(-6x^2))#
#color(red)((du(x))/dx)=-18x^2(-2x^3-3)^2#
#color(blue)(d/dx(cos^-1u(x))=-(d/dx(u(x)))/(sqrt(1-u(x)^2)#
#color(blue)(d/dx(cos^-1u(x))=-(-18x^2(-2x^3-3)^2)/(sqrt(1-((-2x^3-3)^3)^2)#
Therefore,
#color(blue)(d/dx(cos^-1u(x))=(18x^2(-2x^3-3)^2)/(sqrt(1-(-2x^3-3)^6)#
