# What is the derivative of the following funct ion? (1) y=4sin^-1(3×) (2) y=ln(5×+sin^-1(×) (3) y=3cos(sin^-1(2×+3))

Mar 3, 2018

$\left(1\right)$ is $\frac{4}{\sqrt{1 - {x}^{2}}}$

$\left(2\right)$ is $\frac{5 \sqrt{1 - {x}^{2}} + 1}{\left(5 x + {\sin}^{-} 1 \left(x\right)\right) \sqrt{1 - {x}^{2}}}$

$\left(3\right)$ is $- \frac{12 x + 18}{i \sqrt{4 {x}^{2} - 12 x - 10}}$

#### Explanation:

$\left(1\right)$:

$\frac{d}{\mathrm{dx}} \left(4 {\sin}^{-} 1 \left(x\right)\right)$

$4 \cdot \frac{d}{\mathrm{dx}} {\sin}^{-} 1 \left(x\right)$

$4 \cdot \frac{1}{\sqrt{1 - {x}^{2}}}$

$\frac{4}{\sqrt{1 - {x}^{2}}}$

$\left(2\right)$:

$\frac{d}{\mathrm{dx}} \left(\ln \left(5 x + {\sin}^{-} 1 \left(x\right)\right)\right)$

According to the chain rule, $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, where $u$ is a function within $f$.

Here, we have:

$\frac{d}{\mathrm{du}} \ln \left(u\right) \cdot \frac{d}{\mathrm{dx}} \left(5 x + {\sin}^{-} 1 \left(x\right)\right)$

$\frac{1}{u} \cdot \left(5 + \frac{1}{\sqrt{1 - {x}^{2}}}\right)$

But since $u = 5 x + {\sin}^{-} 1 \left(x\right)$, we have:

$\frac{1}{5 x + {\sin}^{-} 1 \left(x\right)} \left(5 + \frac{1}{\sqrt{1 - {x}^{2}}}\right)$

$\frac{5}{5 x + {\sin}^{-} 1 \left(x\right)} + \frac{1}{5 x + {\sin}^{-} 1 \left(x\right) \sqrt{1 - {x}^{2}}}$

$\frac{5 \sqrt{1 - {x}^{2}} + 1}{\left(5 x + {\sin}^{-} 1 \left(x\right)\right) \sqrt{1 - {x}^{2}}}$

$\left(3\right)$:

$\frac{d}{\mathrm{dx}} \left(3 \cos \left({\sin}^{-} 1 \left(2 x + 3\right)\right)\right)$

$3 \cdot \frac{d}{\mathrm{dx}} \cos \left({\sin}^{-} 1 \left(2 x + 3\right)\right)$

Use the chain rule:

$3 \cdot \frac{d}{\mathrm{du}} \cos \left(u\right) \cdot \frac{d}{\mathrm{dx}} {\sin}^{-} 1 \left(2 x + 3\right)$

$3 \cdot \frac{d}{\mathrm{du}} \cos \left(u\right) \cdot \frac{d}{\mathrm{dw}} {\sin}^{-} 1 \left(w\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x + 3\right)$

$3 \cdot \left(- \sin \left(u\right)\right) \cdot \frac{1}{\sqrt{1 - {w}^{2}}} \cdot 2$

$- 6 \cdot \sin \left(u\right) \cdot \frac{1}{\sqrt{1 - {w}^{2}}}$

As $u = {\sin}^{-} 1 \left(2 x + 3\right)$ and $w = 2 x + 3$, we get:

$- 6 \cdot \sin \left({\sin}^{-} 1 \left(2 x + 3\right)\right) \cdot \frac{1}{\sqrt{1 - {\left(2 x + 3\right)}^{2}}}$

$- 6 \cdot \left(2 x + 3\right) \cdot \frac{1}{\sqrt{- 4 {x}^{2} + 12 x + 10}}$

$- \frac{12 x + 18}{i \sqrt{4 {x}^{2} - 12 x - 10}}$

Mar 3, 2018

1) $\frac{12}{\sqrt{1 - 9 {x}^{2}}}$

2)$\frac{5 + \frac{1}{\sqrt{1 - {x}^{2}}}}{5 x + {\sin}^{-} 1 x}$

3) $- \frac{6 \left(2 x + 3\right)}{\sqrt{1 - {\left(2 x + 3\right)}^{2}}}$

#### Explanation:

You can Solve this Using the Chain Rule
basically,

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right)$

$= f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

which is the derivative of the first function($f \left(x\right)$) evaluated at $g \left(x\right)$
multiplied by the derivative of $g \left(x\right)$

therefore for the first question,
let $f \left(x\right) = 4 {\sin}^{-} 1 \left(x\right)$
and $g \left(x\right) = 3 x$

therefore, for the derivative of $f \left(g \left(x\right)\right)$

derivative of f(x) is $4 \cdot \frac{1}{\sqrt{1 - {x}^{2}}}$
and of g(x) Is 3

therefore , the total derivative is

$4 \cdot \frac{1}{\sqrt{1 - {\left(3 x\right)}^{2}}} \cdot 3$
$= \frac{12}{\sqrt{1 - 9 {x}^{2}}}$

2)
similarly, for the second one,
derivative of $\ln \left(x\right)$ is$\frac{1}{x}$ which evaluated at the inside function is $\frac{1}{5 x + {\sin}^{-} 1 \left(x\right)}$ multiplied by the derivative of the inside function of
$5 x + {\sin}^{-} 1 \left(x\right)$
is $5 + \frac{1}{\sqrt{1 - {x}^{2}}}$

therefore the total derivative is
$\frac{5 + \frac{1}{\sqrt{1 - {x}^{2}}}}{5 x + {\sin}^{-} 1 \left(x\right)}$

3)
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} 3 \cos \left({\sin}^{-} 1 \left(2 x + 3\right)\right)$

use the same chain rule twice.
the derivative of the outside function = $- 3 \sin \left(x\right)$
which has to be evaluated at the inside function,
${\sin}^{-} 1 \left(2 x + 3\right)$

And the derivative of the inside function found using the chain rule is
$2 \cdot \frac{1}{\sqrt{1 - {\left(2 x + 3\right)}^{2}}}$

therefore,
$- 3 \sin \left({\sin}^{-} 1 \left(2 x + 3\right)\right) \cdot \frac{2}{\sqrt{1 - {\left(2 x + 3\right)}^{2}}}$

but since $\sin \left({\sin}^{-} 1 \left(2 x + 3\right)\right) = 2 x + 3$
the entire derivative is

$- \frac{6 \left(2 x + 3\right)}{\sqrt{1 - {\left(2 x + 3\right)}^{2}}}$

And That's it