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Answer 1 · 2018-02-06T08:50:15

#f(x)=x^(5/4)# and #g(x)=tanx#

#(fg)(x)# is defined as #(fg)(x)=f(x)g(x)#

Now in #(fg)'(x)=5/4x^(1/4)tanx+x^(5/4)sec^2x#

observe that #5/4x^(1/4)# is diferential of #x^(5/4)#

and #sec^2x# is differential of #tanx#

and it is quite apparent that product rule has been applied on #x^(5/4)tanx#

Hence #(fg)(x)=x^(5/4)tanx#

and we can have #f(x)=x^(5/4)# and #g(x)=tanx#