What are the first and second derivatives of f(x)=2ln(sqrt((4x-4)/(6x+10))) ?

1 Answer
Dec 25, 2015

We'll need the chain rule and the quotient rule here.

Explanation:

  • Chain rule states that (dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)

  • Quotient rule states that for y=f(x)/g(x), then y'=(f'g-fg')/g^2

Renaming u=sqrt(v) and v=(4x-4)/(6x+10), we have the new f(x)=2ln(u) and can now start, but let's do it step-by-step.

(dy)/(du)=2/u

(du)/(dv)=1/(2sqrt(v))

(dv)/(dx)=((4)(6x+10)-(4x-4)(6))/(6x+10)^2=(24x+40-24x+24)/(6x+10)^2=64/(6x+10)^2

Aggregating:

(dy)/(dx)=128/(u*2sqrt(v)*(6x+10)^2)

Substituting u:

(dy)/(dx)=128/(sqrt(v)2sqrt(v)(6x+10)^2)=128/(2v*(6x+10)^2)

Substituting v (Holy Jesus!):

(dy)/(dx)=128/(2((4x-4)/(cancel(6x+10)))(6x+10)^(cancel(2)))

(dy)/(dx)=64/((4x-4)(6x+10))=64/(24x^2+16x-40)

(dy)/(dx)=8/(3x^2+2x-5)

Now, the second derivative:

(dy^2)/(d^2x)=((0)(3x^2+2x+5)-(8)(6x+2))/(3x^2+2x-5)^2=-(48x+16)/(3x^2+2x-5)^2