How do you use the chain rule to differentiate #y=((x^5+4)/(x^2-5))^(1/5)#?

1 Answer
Noah G
Mar 13, 2018

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

Explanation:

I wouldn't use the chain rule, instead I would use logarithmic differentiation.

#lny = ln((x^5 + 4)/(x^2 - 5))^(1/5)#

#lny = 1/5ln((x^5 + 4)/(x^2 -5))#

#lny = 1/5(ln(x^5 + 4) - ln(x^2 - 5))#

#lny = 1/5ln(x^5 + 4) - 1/5ln(x^2 - 5)#

#1/y(dy/dx) = 1/5(5x^4)/(x^5 + 4) - 1/5(2x)/(x^2 - 5)#

#1/y(dy/dx) = x^4/(x^5 + 4) - (2x)/(5(x^2 - 5))#

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

Hopefully this helps!

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