find $\frac{dy}{dx}$ where $y = sin^{-1}(\frac{1}{2x + 5})$. provide your answer below.

Explanation:

Step1: Recall derivative of inverse - sine function

The derivative of $y = \sin^{-1}(u)$ with respect to $x$ is $\frac{dy}{dx}=\frac{1}{\sqrt{1 - u^{2}}}\cdot\frac{du}{dx}$ by the chain - rule. Here $u=\frac{1}{2x + 5}$.

Step2: Find the derivative of $u$ with respect to $x$

Using the quotient rule, if $u=\frac{1}{2x + 5}=(2x + 5)^{-1}$, then $\frac{du}{dx}=-1\times(2x + 5)^{-2}\times2=-\frac{2}{(2x + 5)^{2}}$.

Step3: Substitute $u$ and $\frac{du}{dx}$ into the chain - rule formula

First, $1 - u^{2}=1-\left(\frac{1}{2x + 5}\right)^{2}=\frac{(2x + 5)^{2}-1}{(2x + 5)^{2}}=\frac{4x^{2}+20x + 25 - 1}{(2x + 5)^{2}}=\frac{4x^{2}+20x + 24}{(2x + 5)^{2}}$. Then $\sqrt{1 - u^{2}}=\frac{\sqrt{4x^{2}+20x + 24}}{|2x + 5|}$. $\frac{dy}{dx}=\frac{1}{\sqrt{1 - u^{2}}}\cdot\frac{du}{dx}=\frac{1}{\frac{\sqrt{4x^{2}+20x + 24}}{|2x + 5|}}\cdot\left(-\frac{2}{(2x + 5)^{2}}\right)=-\frac{2}{(2x + 5)\sqrt{4x^{2}+20x + 24}}=-\frac{2}{2(2x + 5)\sqrt{x^{2}+5x + 6}}=-\frac{1}{(2x + 5)\sqrt{x^{2}+5x + 6}}$.

Answer:

$-\frac{1}{(2x + 5)\sqrt{x^{2}+5x + 6}}$