How do you use the chain rule to differentiate #root9(e^(6x))#?

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Answer 1 · 2017-03-16T17:02:31

#f(x)=2/3root(3)(e^(2x))#

According to chain rule when #f=f(g(h(x)))#,

#(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)#

Here #f(x)=root(9)g(x)#, #g(x)=e^(h(x))# and #h(x)=6x#

and #(df)/(dg)=1/9(g(x)^((1/9-1)))#

#(dg)/(dx)=e^(h(x))# and #(dh)/(dx)=6#

Hence as #f(x)=root(9)(e^(6x))#

#(df)/(dx)=1/9(g(x)^((1/9-1)))xxe^(h(x))xx6#

= #6/9(e^(6x))^((1/9-1))e^(6x)#

= #6/9e^(6x xx(1/9-1))e^(6x)#

= #6/9e^(6x xx(1/9-1)+6x)#

= #6/9e^((6x)/9)#

= #2/3root(3)(e^(2x))#