How do you use the chain rule to differentiate #y=(2x-1)^4/(x+1)^2#?

1 Answer
Jim G.
Dec 20, 2017

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

Explanation:

#"using a combination of "color(blue)"chain/quotient rule"#

#"given "y=(g(x))/(h(x))" then"#

#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#g(x)=(2x-1)^4rArrg'(x)=4(2x-1)^3xxd/dx(2x-1)#

#color(white)(xxxxxxxxxxxxxxxxxx)=8(2x-1)^3#

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

#dy/dx=((x+1)^2 .8(2x-1)^3-(2x-1)^4 .2(x+1))/(x+1)^4#

#color(white)(dy/dx)=(2(x+1)(2x-1)^3[4(x+1)-(2x-1)])/(x+1)^4#

#color(white)(dy/dx)=(2(x+1)(2x-1)^3(2x+5))/(x+1)^4#

#color(white)(dy/dx)=(2(2x-1)^3(2x+5))/(x+1)^3#

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