# Question 2cc56

Apr 5, 2017

Use the product rule and the chain rule.

#### Explanation:

Given: $f \left(x\right) = x {\sec}^{-} 1 \left({x}^{3}\right)$

The product rule is:

$\frac{d}{\mathrm{dx}} \left(u \cdot v\right) = \frac{\mathrm{du}}{\mathrm{dx}} \cdot v + u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$

For the given function:

Let $u = x$ and $v = {\sec}^{-} 1 \left({x}^{3}\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

The computation of $\frac{\mathrm{dv}}{\mathrm{dx}}$ requires the use of the chain rule

$\frac{d \left(g \left(h \left(x\right)\right)\right)}{\mathrm{dx}} = \frac{\mathrm{dg}}{\mathrm{dh}} \frac{\mathrm{dh}}{\mathrm{dx}}$

Let $h \left(x\right) = {x}^{3}$ and $g = {\sec}^{-} 1 \left(h\right)$

$\frac{\mathrm{dh}}{\mathrm{dx}} = 3 {x}^{2}$

(dg)/(dh) = 1/(h^2sqrt(1 - 1/h^2)#

Substitute these into the right side of the chain rule:

$\frac{d \left(g \left(h \left(x\right)\right)\right)}{\mathrm{dx}} = \left(\frac{1}{{h}^{2} \sqrt{1 - \frac{1}{h} ^ 2}}\right) \left(3 {x}^{2}\right)$

Reverse the substitution for h:

$\frac{d \left(g \left(h \left(x\right)\right)\right)}{\mathrm{dx}} = \left(\frac{1}{{\left({x}^{3}\right)}^{2} \sqrt{1 - \frac{1}{{x}^{3}} ^ 2}}\right) \left(3 {x}^{2}\right)$

Two powers of x cancel and the 3 can move to the numerator:

$\frac{d \left(g \left(h \left(x\right)\right)\right)}{\mathrm{dx}} = \frac{3}{\left({x}^{4}\right) \sqrt{1 - \frac{1}{{x}^{6}}}}$

Returning to the product rule:

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{3}{\left({x}^{4}\right) \sqrt{1 - \frac{1}{{x}^{6}}}}$

Substitute the values into the product rule:

$\frac{d}{\mathrm{dx}} \left(x {\sec}^{-} 1 \left({x}^{3}\right)\right) = {\sec}^{-} 1 \left({x}^{3}\right) + x \cdot \left(\frac{3}{\left({x}^{4}\right) \sqrt{1 - \frac{1}{{x}^{6}}}}\right)$

one power of x cancels in the second term:

$\frac{d}{\mathrm{dx}} \left(x {\sec}^{-} 1 \left({x}^{3}\right)\right) = {\sec}^{-} 1 \left({x}^{3}\right) + \frac{3}{\left({x}^{3}\right) \sqrt{1 - \frac{1}{{x}^{6}}}}$

Apr 5, 2017

#### Explanation:

We know that differential of inverse trigonometric ratio
$g \left(x\right) = {\sec}^{-} 1 \left(x\right)$
$g ' \left(x\right) = \frac{1}{x \sqrt{{x}^{2} - 1}}$

$\therefore$ if $g \left(x\right) = {\sec}^{-} 1 \left({x}^{3}\right)$
We get $g ' \left(x\right) = \frac{1}{{x}^{3} \sqrt{{\left({x}^{3}\right)}^{2} - 1}}$ ........(1)

Given expression is
$f \left(x\right) = x {\sec}^{-} 1 \left({x}^{3}\right)$
Using chain rule we get
$f ' \left(x\right) = x \frac{d}{\mathrm{dx}} \left({\sec}^{-} 1 \left({x}^{3}\right)\right) + \left({\sec}^{-} 1 \left({x}^{3}\right)\right) \times \frac{d}{\mathrm{dx}} x$
Using (1) we get
$f ' \left(x\right) = x \times \left(3 {x}^{2}\right) \times \left(\frac{1}{{x}^{3} \sqrt{{\left({x}^{3}\right)}^{2} - 1}}\right) + \left({\sec}^{-} 1 \left({x}^{3}\right)\right) \times 1$
$\implies f ' \left(x\right) = \frac{3 {x}^{3}}{{x}^{3} \sqrt{{x}^{6} - 1}} + {\sec}^{-} 1 \left({x}^{3}\right)$
$\implies f ' \left(x\right) = \frac{3}{\sqrt{{x}^{6} - 1}} + {\sec}^{-} 1 \left({x}^{3}\right)$

Apr 5, 2017

#### Explanation:

make use of this to write for $\tan \left(\frac{y}{x}\right) = \tan \left({\sec}^{-} 1 \left({x}^{3}\right)\right) = \sqrt{{x}^{6} - 1}$