Question

find the derivative of y with respect to x.
y = \csc^{-1}(6x^{2}+1), x > 0
\frac{dy}{dx}=\square

Explanation:

Step1: Recall the derivative formula for $y = \csc^{-1}(u)$

The derivative of $y=\csc^{-1}(u)$ with respect to $x$ is $\frac{dy}{dx}=-\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{dx}$ when $|u| > 1$. Here $u = 6x^{2}+1$, and since $x>0$, $u=6x^{2}+1>1$.

Step2: Find $\frac{du}{dx}$

Differentiate $u = 6x^{2}+1$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $\frac{du}{dx}=\frac{d}{dx}(6x^{2}+1)=12x$.

Step3: Substitute $u$ and $\frac{du}{dx}$ into the derivative formula

Substitute $u = 6x^{2}+1$ and $\frac{du}{dx}=12x$ into $\frac{dy}{dx}=-\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{dx}$. Since $u = 6x^{2}+1>0$ for $x > 0$, $|u|=u$. So $\frac{dy}{dx}=-\frac{12x}{(6x^{2}+1)\sqrt{(6x^{2}+1)^{2}-1}}$.

Answer:

$-\frac{12x}{(6x^{2}+1)\sqrt{(6x^{2}+1)^{2}-1}}$