Find the derivative of each function: y = 1/root x, y = x/2 - 2/x, y = x times (root x minus x), and y = (3x - 2)(3x + 2).

Step-by-step solution

1. 1

   Notice what these four have in common. None of them is a plain sum of powers as written — there are roots, quotients and products. The power rule $\dfrac{d}{dx}x^{n}=nx^{n-1}$ applies only to powers, so each one has to be rewritten first. Doing that is almost always faster than reaching for the quotient or product rule.

2. 2

   (1) Rewrite the root as a negative fractional power.

   $y=\frac{1}{\sqrt{x}}=x^{-1/2}\ \Longrightarrow\ y'=-\tfrac12 x^{-3/2}=-\frac{1}{2x^{3/2}}=-\frac{1}{2x\sqrt{x}}.$

3. 3

   (2) Split the quotient into two simple terms.

   $y=\tfrac12 x-2x^{-1}\ \Longrightarrow\ y'=\tfrac12-2(-1)x^{-2}=\frac{1}{2}+\frac{2}{x^{2}}.$

   The sign flip on the second term is the usual trap: differentiating $-2x^{-1}$ gives $+2x^{-2}$.

4. 4

   (3) Expand the product before differentiating.

   $y=x\left(x^{1/2}-x\right)=x^{3/2}-x^{2}\ \Longrightarrow\ y'=\tfrac32 x^{1/2}-2x=\frac{3\sqrt{x}}{2}-2x.$

5. 5

   (4) Multiply out the conjugate pair. It is a difference of squares, so the middle terms cancel:

   $y=(3x)^{2}-2^{2}=9x^{2}-4\ \Longrightarrow\ y'=18x.$

   The constant $-4$ differentiates to $0$.

6. 6

   Collect the four answers.

   $y'=-\frac{1}{2x^{3/2}},\qquad y'=\frac{1}{2}+\frac{2}{x^{2}},\qquad y'=\frac{3\sqrt{x}}{2}-2x,\qquad y'=18x.$

   For (1) and (3) the domain is $x>0$; for (2) it is $x\neq 0$. Each result was confirmed by symbolic differentiation of the original unrewritten form.