What is the derivative of #y= (5x)/sqrt (x^2+9)#?

Question

2162 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T15:43:24

#y'=45/(x^2+9)^(3/2)#

By rewriting a bit,

#y=5 x/(x^2+9)^(1/2)#

#y'=5(1 cdot (x^2+9)^(1/2)-x cdot1/(cancel2)(x^2+9)^(-1/2)(cancel 2x))/((x^2+9)^(1/2))^2#

By cleaning up a bit,

#=5((x^2+9)^(1/2)-x^2/(x^2+9)^(1/2))/(x^2+9)#

By multiply the numerator and the denominator by #(x^2+9)^(1/2)#,

#=5(cancel(x^2)+9-cancel(x^2))/(x^2+9)^(3/2)=45/(x^2+9)^(3/2)#

I hope that this was clear.