Question

question. find $\frac{dy}{dx}$ where $y = csc^{-1}(2x^{4}+6)$. provide your answer below. $\frac{dy}{dx}=square$

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 $u = 2x^{4}+6$.

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

If $u = 2x^{4}+6$, then by the power - rule $\frac{du}{dx}=\frac{d}{dx}(2x^{4}+6)=8x^{3}$.

Step3: Apply the chain - rule

Substitute $u = 2x^{4}+6$ and $\frac{du}{dx}=8x^{3}$ into the formula for $\frac{dy}{dx}$ of $y=\csc^{-1}(u)$.
$\frac{dy}{dx}=-\frac{8x^{3}}{|2x^{4}+6|\sqrt{(2x^{4}+6)^{2}-1}}$.

Answer:

$-\frac{8x^{3}}{|2x^{4}+6|\sqrt{(2x^{4}+6)^{2}-1}}$