How do you find the derivative of $y = \sqrt{3 x + 1}$ $y =sqrt(3x+1)$ ?

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Answer 1 · 2014-07-31T14:30:54

When you're differentiating radicals, the key is to rewrite them in rational exponent form:

$y = {(3 x + 1)}^{\frac{1}{2}}$ $y = (3x+1)^(1/2)$

Now it's more clear that we can apply the power rule and then the chain rule to find this function's derivative:

$\frac{d y}{d x} = 3 \cdot (\frac{1}{2} \cdot {(3 x + 1)}^{- \frac{1}{2}})$ $dy/dx = 3*(1/2*(3x+1)^(-1/2))$

And now all we need to do is simplify a bit:

$\frac{d y}{d x} = \frac{3}{2 \sqrt{3 x + 1}}$ $dy/dx = 3/(2sqrt(3x+1))$

How do you find the derivative of y=√3x+1y =sqrt(3x+1)?