Question

find the derivative of y with respect to x.
y = csc^(-1)(4x^2 + 1), x > 0
dy/dx = □

Explanation:

Step1: Recall derivative formula

The derivative of $y = \csc^{-1}(u)$ with respect to $x$ is $\frac{dy}{dx}=-\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{dx}$ (by the chain - rule), where $u = 4x^{2}+1$.

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

Since $u = 4x^{2}+1$, then $\frac{du}{dx}=\frac{d}{dx}(4x^{2}+1)=8x$.

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

Since $x>0$, then $u = 4x^{2}+1>0$. Substitute $u = 4x^{2}+1$ and $\frac{du}{dx}=8x$ into $\frac{dy}{dx}=-\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{dx}$. We get $\frac{dy}{dx}=-\frac{8x}{(4x^{2}+1)\sqrt{(4x^{2}+1)^{2}-1}}$.

Answer:

$-\frac{8x}{(4x^{2}+1)\sqrt{(4x^{2}+1)^{2}-1}}$