# `y=sqrt(3x)+sqrt(5x)` y=sqrt over 3x+sqrt 5x (the sq rt over 3 extends over the whole equation and the plus sqrt over 5x is all under that sqrt 3

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### 1 Answer

Supposing that you need to differentiate the function `y = sqrt(3x + sqrt(5x)) ` with respect to x, you need to use chain rule, such that:

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

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

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

**Hence, evaluating the derivative of the given function `y = sqrt(3x + sqrt(5x))` yields **`(dy)/(dx) = (3 + 5/(2sqrt(5x)))/(2sqrt(3x + sqrt(5x))).`