# y=sqrt 3x +sqrt5x, dy/dx=? The square root extends over the top of the 5x square root and I can not figure out how to solve it

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`y = sqrt3x+sqrt5x`

We know that if;

`y = sqrt(ax) ` then

`dy/dx =1/(2sqrt(ax))xxa`

Similarly;

`dy/dx = 1/(2sqrt(3x))xx3+1/(2sqrt(5x))xx5`

`dy/dx =3/(2sqrt(3x))+5/(2sqrt(5x))`

**So the answer is;**

`dy/dx =3/(2sqrt(3x))+5/(2sqrt(5x))`

`y=sqrt(3x)+sqrt(5x)`

`dy/dx=(1/2)1/(sqrt(3x))(3)+(1/2)(1/(sqrt(5x))(5)`

`dy/dx=(1/(2sqrt(x)))(sqrt(3)+sqrt(5))`