# How do you find the derivative of sin^-1(2x+1)?

Nov 2, 2016

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

#### Explanation:

For this equation you would use the [chain rule] (https://socratic.org/calculus/basic-differentiation-rules/chain-rule) so you take the derivative of the outside:
$\left({\sin}^{-} 1\right)$
times the derivative of the inside:
$\left(2 x + 1\right)$

So the derivative of ${\sin}^{-} 1$ otherwise known as $\arcsin$ is $\frac{1}{\sqrt{1 - {x}^{2}}}$
Differentiating Inverse Trigonometric Functions

but in this case $\left(2 x - 1\right)$ is acting as $x$ so it's
$\frac{1}{\sqrt{1 - {\left(2 x - 1\right)}^{2}}}$

Next the derivative of $2 x - 1$ is $2$

So the answer becomes outside times inside
Which is

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

Here are the derivatives of inverse functions

Nov 4, 2016

$\textcolor{g r e e n}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\sqrt{1 - {\left(2 x + 1\right)}^{2}}}}$

#### Explanation:

Given:$\text{ Determine } \frac{d}{\mathrm{dx}} \left[{\sin}^{- 1} \left(2 x + 1\right)\right]$

$\textcolor{b l u e}{\text{Preamble}}$

Note that ${\sin}^{- 1}$ has a particular meaning which has the alternative presentation of Arcsin. It is not connected to the form example of ${\sin}^{2}$ which is sin squared

So ${\sin}^{- 1} \left(2 x + 1\right) \to \arcsin \left(2 x + 1\right)$

Note that ${\sin}^{- 1} \left(2 x + 1\right)$ is another way of writing an angle and the sin of which gives the value $2 x + 1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Answering the question}}$

Set $\theta = {\sin}^{- 1} \left(2 x + 1\right)$

Note that $\sin \left(\theta\right) = \left(\text{Opposite")/("Hypotenus}\right)$

Set the value of the Hypotenuse to 1 giving:

$\sin \left(\theta\right) = \left(\text{Opposite")/("Hypotenuse}\right) = \frac{2 x + 1}{1}$

Using $\sin \left(\theta\right) = 2 x + 1$ and implicitly differentiating

$\cos \left(\theta\right) \times \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\cos} \left(\theta\right)$.......................Equation(1)

But $\text{Hypotenuse "xxcos(theta) =" Adjacent } = x$

as the hypotenuse is 1 we have $\cos \left(\theta\right) = x$

So equation(1) becomes:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x} \text{ .......................Equation} \left({1}_{a}\right)$

But by Pythagoras ${x}^{2} = {1}^{2} - {\left(2 x + 1\right)}^{2} \text{ "=>" } x = \sqrt{1 - {\left(2 x + 1\right)}^{2}}$

Thus $\text{Equation} \left({1}_{a}\right)$ becomes

$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\sqrt{1 - {\left(2 x + 1\right)}^{2}}}}$