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)
