- Chain rule: #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#
Where #f(x)=u^(3/2)#, #u=ln(v)# and #v=1-x#.
#(df(x))/(dx)=(3u)^(1/2)/2(1/v)(-1)#
Substituting #u#:
#(df(x))/(dx)=-(3(ln(v))^(1/2))/(2v)#
Substituting #v#:
#(df(x))/(dx)=color(green)((3sqrt(ln(1-x)))/(2(x-1)))#
The second derivative falls into the same logic for chain rule in the numerator. However, we have a fraction, which will demand quotient rule.
- Quotient rule: be #y=f(x)/g(x)#, then #y'=(f'g-fg')/g^2#
#(df(x)^2)/(d^2x)=(((3/2u^(-1/2))(1/v)(-1))(2(x-1))-3sqrt(ln(1-x))(2)}/(4(x-1)^2)#
Substituting #u#:
#(df(x)^2)/(d^2x)=(((3/(2sqrt(ln(v)))(1/v)(-1))(2(x-1))-6sqrt(ln(1-x))))/(4(x-1)^2)#
Substituting #v#:
#(df(x)^2)/(d^2x)=(((3/(2sqrt(ln(1-x)))(1/(1-x))(-1))(2(x-1))-6sqrt(ln(1-x))))/(4(x-1)^2)#
#(df(x)^2)/(d^2x)=((cancel(-)6cancel((x-1)))/(2cancel((1-x))sqrt(ln(x-1)))-6sqrt(ln(1-x)))/(4(x-1)^2)#
#(df(x)^2)/(d^2x)=((3-6ln(1-x))/(sqrt(ln(1-x))))/(4(x-1)^2)=color(green)((3-6ln(1-x))/(4(x-1)^2sqrt(ln(1-x))))#
