How do you differentiate # f(x)=sqrt((ln(-x+3))^3# using the chain rule.?

1 Answer
Shwetank Mauria
May 8, 2017

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

Explanation:

Chain Rule - In order to differentiate a function of a function, say #y, =f(g(x))#, where we have to find #(dy)/(dx)#, we need to do (a) substitute #u=g(x)#, which gives us #y=f(u)#. Then we need to use a formula called Chain Rule, which states that #(dy)/(dx)=(dy)/(du)xx(du)/(dx)#. In fact if we have something like #y=f(g(h(k(x))))#, we can have #(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)xx(dk)/(dx)#

Here we have #f(x)=sqrt((ln(-x+3))^3)# or say

#f(x)=sqrt(g(x))#, where #g(x)=(h(x))^3#, #h(x)=ln(k(x))# and #k(x)=-x+3#

and #(df)/(dg)=1/(2sqrt(g(x))#, #(dg)/(dh)=3(h(x)^2)#, #(dh)/(dk)=1/(k(x))# and #(dk)/(dx)=-1#

hence #(df)/(dx)=1/(2sqrt(g(x))xx3(h(x)^2)xx1/(k(x))xx(-1)#

= #1/(2sqrt((ln(-x+3))^3))xx3(ln(-x+3))^2xx1/((-x+3))xx(-1)#

= #-(3(ln(-x+3))^2)/(2(-x+3)sqrt((ln(-x+3))^3)#

= #-(3(ln(3-x))^2)/(2(3-x)sqrt((ln(3-x))^3)#

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