Question

How do you find the derivative with a square root in the denominator `y= 5x/sqrt(x^2+9)`?

Answer 1

`dy/dx=(10(x^2+9)-5x)/(sqrt(x^2+9)(2x^2+18))`

Explanation:

The Quotient Rule says

`d/dx(f(x)/g(x))=(g(x)d/dx(f(x))-f(x)d/dx(g(x)))/(g(x))^2`

If we let `f(x) = 5x` and `g(x)=sqrt(x^2+9)`, so that
`f'(x)=5` and `g'(x)=1/2(x^2+9)^(-1/2)`, and
`(g(x))^2=x^2+9`. Plugging these results in the Quotient Rule, we then have

`dy/dx=d/dx(f(x)/g(x))=(5sqrt(x^2+9)-5x1/2(x^2+9)^(-1/2))/(x^2+9)`
`=((2 xx)/(2 xx))(5sqrt(x^2+9)-5x1/2(x^2+9)^(-1/2))/(x^2+9)`
`=((sqrt(x^2+9) xx)/(sqrt(x^2+9) xx))(10sqrt(x^2+9)-5x(x^2+9)^(-1/2))/(2x^2+18)`
`=(10(x^2+9)-5x)/(sqrt(x^2+9)(2x^2+18))`