How do you differentiate #f(x)= (x-1)/( x+3) ^ (1/3 # using the quotient rule?

1 Answer
Jim G.
May 1, 2017

#f'(x)=(2x+4)/(3(x+3)^(4/3))#

Explanation:

#"Given " f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"#

#"also require to differentiate using the "color(blue)"chain rule"#

#d/dx(f(g(x)))=f'(g(x)xxg'(x)#

#"here " g(x)=x-1rArrg'(x)=1#

#h(x)=(x+3)^(1/3)rArrh'(x)=1/3(x+3)^(-2/3)#

#rArrf'(x)=((x+3)^(1/3)-(x-1) . 1/3(x+3)^(-2/3))/(x+3)^(2/3)#

#=(1/3(x+3)^(-2/3)[3(x+3)-(x-1)])/(x+3)^(2/3)#

#=(2x+4)/(3(x+3)^(4/3))#

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