How do you differentiate #f(x)=x(1-2x+x^2)^(3/5)# using the chain rule?

1 Answer
Dec 24, 2015

Answer:

Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#.

Explanation:

Renaming #u=1-2x+x^2#, we have now #f(x)=x*u^(3/5)#

If we could not derivate directly #(1-3x+x^2)^(3/5)#, now we can do so using chain rule. Also, we see that this is all about a product (first term is #x# and the second is #u^(3/5)#).

To derivate a product, we must remember the product rule, which states that, for a function #y=f(x)g(x)#, #(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#

So:

#(df(x))/(dx)=1*u^(3/5)+x*(3u^(-2/5))/5#

#(df(x))/(dx)=(1-2x+x^2)^(3/5)+(3x)/(5(1-2x+x^2)^(2/5))#

#(df(x))/(dx)=(5(1-2x+x^2)+3x)/(5(1-2x+x^2)^(2/5)#