How do you differentiate y =1/sqrtln(x^2-3x) using the chain rule?

1 Answer
Dec 31, 2015

We can rewrite it as y=(ln(x^2-3x))^(-1/2) and use the chain rule, which states that (dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)

Explanation:

Renaming:

y=u^(-1/2)
u=ln(v)
v=x^2-3x

(dy)/(dx)=-1/(2u^(3/2))*1/v(2x-3)

(dy)/(dx)=-(2x-3)/(2u^(3/2)v)

Substituting u:

(dy)/(dx)=-(2x-3)/(2(ln(v))^(3/2)*v)

Substituting v:

(dy)/(dx)=-(2x-3)/(2(ln(x^2-3x))^(3/2)*(x^2-3x))

Finally:

(dy)/(dx)=(3-2x)/(2(x^2-3x)sqrt((ln(x^2-3x))^3