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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jeew-m's profile pic

Posted on (Answer #1)

`y = sqrt3x+sqrt5x`


We know that if;

`y = sqrt(ax) ` then

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



`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))`


pramodpandey's profile pic

Posted on (Answer #2)




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