1. ## Derivatives help

Hi, I'm having difficulty solving and simplifying these two derivatives:

y = x
_____
√(x^2-1)

and

y= (3x^2 - 2)^2
√(x^2-5)

Thank you for any help!

2. You can try the product rule on the first one.

$x(x^{2}-1)^{\frac{-1}{2}}$

$x(\frac{-1}{2})(x^{2}-1)^{\frac{-3}{2}}(2x)+(x^{2}-1)^{\frac{-1}{2}}$

$-x^{2}(x^{2}-1)^{\frac{-3}{2}}+(x^{2}-1)^{\frac{-1}{2}}$

Factor:

$(x^{2}-1)^{\frac{-1}{2}}(1-x^{2}(x^{2}-1)^{-1})$

$(x^{2}-1)^{\frac{-1}{2}}(\frac{-1}{x^{2}-1})$

$\frac{-1}{(x^{2}-1)^{\frac{3}{2}}}$

3. Hello, starswept!

There are some tricks to simplifying expressions with negative exponents.

$y \;=\;\frac{x}{\sqrt{x^2-1}}$
We have: . $y \;=\;\frac{x}{(x^2-1)^{\frac{1}{2}}}$

Quotient Rule: . $\frac{dy}{dx} \;=\;\frac{(x^2-1)^{\frac{1}{2}}\cdot1 - x\cdot\frac{1}{2}(x^2-1)^{-\frac{1}{2}}\cdot2x}{\left[(x^2-1)^{\frac{1}{2}}\right]^2}\;=\;\frac{(x^2-1)^{\frac{1}{2}} - x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1}$

Multiply by $\frac{(x^2-1)^{\frac{1}{2}}}{(x^2-1)^{\frac{1}{2}}}$

. . $\frac{dy}{dx}\;=\;\frac{(x^2-1)^{\frac{1}{2}}}{(x^2-1)^{\frac{1}{2}}}\cdot\frac{(x^2-1)^{\frac{1}{2}} - x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1} \;= \;\frac{x^2 - 1 - x^2}{(x^2-1)^{\frac{3}{2}}}$

Therefore: . $\frac{dy}{dx}\;=\;\frac{-1}{(x^2-1)^{\frac{3}{2}}}$

$y \;= \;(3x^2 - 2)^2\sqrt{x^2-5}$

We have: . $y \;=\;(3x^2-2)^2(x^2-1)^{\frac{1}{2}}$

Produce Rule: . $\frac{dy}{dx} \;= \;(3x^2-2)^2\cdot\frac{1}{2}(x^2-5)^{-\frac{1}{2}}\cdot2x \:+ \:(x^2-5)^{\frac{1}{2}}\cdot2(3x^2-2)\cdot6x$

. . . . . . . . . . . $\frac{dy}{dx}\;= \;x(3x^2-2)^2(x^2-5)^{-\frac{1}{2}} + 12x(3x^2-2)(x^2-5)^{\frac{1}{2}}$

Factor: . $\frac{dy}{dx} \;=\;x(3x^2-2)\,\left[(3x^2-2)(x^2-5)^{-\frac{1}{2}} + 12(x^2-5)^{\frac{1}{2}}\right]$

Multiply by $\frac{(x^2-5)^{\frac{1}{2}}}{(x^2-5)^{\frac{1}{2}}}$

. . $\frac{dy}{dx} \;= \;x(3x^2-2)\!\cdot\!\frac{(3x^2-2)(x^2-5)^{-\frac{1}{2}} + 12(x^2-5)^{\frac{1}{2}}}{1}\cdot\frac{(x^2-5)^{\frac{1}{2}}}{(x^2-5)^{\frac{1}{2}}}$
.
. . $\frac{dy}{dx} \;=\;x(3x^2-2)\cdot\frac{(3x^2 - 2) + 12(x^2-5)}{(x^2-5)^{\frac{1}{2}}}
$

Therefore: . $\frac{dy}{dx}\;=\;\frac{x(3x^2-2)(15x^2-62)}{(x^2-5)^{\frac{1}{2}}}$