# finding the derivative of a square root function

I need to find $y'$ if $y=\sqrt x(x-1)$

I distributed the sqrt of x: $\sqrt x(x)-\sqrt x$

what do I do now? I need to get rid of the square root in order to find the derivative but I'm not sure how.

Hint: Observe that: $$y = \sqrt{x}(x - 1) = x^{1/2}(x^{2/2} - 1) = x^{3/2} - x^{1/2}$$

• I get where x^3/2 comes from, but why is there the "-x"? if you're just distributing the x^1/2, wouldn't it be "x^3/2 - x^1/2"?
– lisa
Commented Oct 2, 2013 at 1:30
• Yes, my mistake. Commented Oct 2, 2013 at 1:42

$$y=x\sqrt x-\sqrt x$$ $$=x(x^{1 \over 2})-x^{1 \over 2}$$ $$=x^{3 \over2} - x^{\frac 12}$$ So $$\frac{d}{dx}\left(x^{3 \over2} - x^{\frac 12}\right) = \dfrac 32x^{\frac 12}-\dfrac 12x^{-\frac 12}$$ $$=\dfrac {3\sqrt x}{2}-\dfrac 1{2\sqrt x}$$ We are technically done, but lets rationalise the denominator $$\frac{d}{dx}\left(x^{3 \over2} - x^{\frac 12}\right) = \dfrac {3\sqrt x}{2}-\dfrac{\sqrt x}{2x}$$

First, $(\sqrt{x})' =(x^{1/2})' =(1/2)x^{-1/2}$ since $(x^a)' =a x^{a-1}$.

By the product rule

\begin{align} (\sqrt{x}(x-1))' &=\sqrt{x}(x-1)' + (\sqrt{x})'(x-1)\\ &=\sqrt{x}+(1/2)x^{-1/2}(x-1)\\ &=\sqrt{x}+(1/2)(x^{1/2}-x^{-1/2})\\ &=\sqrt{x}+(1/2)\sqrt{x}-1/(2\sqrt{x})\\ &=(3/2)\sqrt{x}-1/(2\sqrt{x})\\ &=(3/2)\sqrt{x}-\sqrt{x}/(2x)\\ &=\frac{\sqrt{x}}{2}\left(3-\frac{1}{2x}\right)\\ \end{align}

Those last three lines are all, in my opinion, acceptable ways to wrote the result.