# What is the derivative of `y = x/sqrt(x^3+1)` ?

### 4 Answers | Add Yours

It is given that y = `x / (sqrt(x^3+1))`

Use the quotient rule and the chain rule

y' = `((x)'*sqrt(x^3+1) - x*(sqrt(x^3+1))')/(x^3 +1)`

=> `(sqrt(x^3+1) - x*(1/2)*(1/(sqrt(x^3+1))*3x^2))/(x^3 +1)`

=> `(1 - x^3/2)/(x^3 +1)^(3/2)`

**The required derivative is** `(1 - x^3/2)/(x^3 +1)^(3/2)`

since for all functions

f=x

g=`1/sqrt(x^3+1)`

=> `g=(x^3+1)^(-1/2)

f'=1

`g'=-(1/2)*(x^3+1)^(-1/2-1)*3x^2`

so `dy/dx=f'g+fg'`

`dy/dx=1/sqrt(x^3+1)+x*(-1/2)*(x^3+1)^(-3/2)*3x^2`

=> `dy/dx=1/sqrt(x^3+1)-(3x^3)/(2sqrt(x^3+1))`

set

f=x

g=

that is

f'=1

since

=> `dy/dx=1/sqrt(x^3+1)-(3x^3)/(2sqrt((x^3+1)^3))

I am not sure if this answer is correct...

y=x(x^3+1)^(-1/2)

.....

y= (-x^3+2) / 2 (sqrt(x^3+1)^3)