Find the derivative of y with respect to x.
-$$y = - \csc ^ { - 1 } \left( \frac { 8 x + 9 } { 3 } \right)$$
A) $\frac { 24 } { \sqrt { ( 8 x + 9 ) ^ { 2 } - 9 } }$
B) $\frac { 24 } { ( 8 x + 9 ) \sqrt { ( 8 x + 9 ) ^ { 2 } - 9 } }$
C) $\frac { - 24 } { 1 + ( 8 x + 9 ) ^ { 2 } }$
D) $\frac { - 24 } { ( 8 x + 9 ) \sqrt { ( 8 x + 9 ) ^ { 2 } - 1 } }$

Question

Quizplus · Accepted Answer

The derivative of $y = -\csc^{-1}\left(\frac{8x + 9}{3}\right)$ can be found using the chain rule and the derivative formula for $\csc^{-1}x$, which is $\frac{-1}{|x|\sqrt{x^2 - 1}}$. Let $u = \frac{8x + 9}{3}$, then $y = -\csc^{-1}(u)$ and $\frac{du}{dx} = \frac{8}{3}$. Applying the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$, gives $\frac{-1}{|u|\sqrt{u^2 - 1}} \cdot \frac{8}{3}$. Substituting $u = \frac{8x + 9}{3}$ back in, we get $\frac{-8}{3} \cdot \frac{1}{\left|\frac{8x + 9}{3}\right|\sqrt{\left(\frac{8x + 9}{3}\right)^2 - 1}}$, which simplifies to $\frac{-24}{(8x + 9)\sqrt{(8x + 9)^2 - 9}}$, taking into account the absolute value and the negative sign from the derivative of $\csc^{-1}x$.

Find the Derivative of Y with Respect to X y=−csc⁡−1(8x+93)y = - \csc ^ { - 1 } \left( \frac { 8 x + 9 } { 3 } \right)y=−csc−1(38x+9​)

Find the Derivative of Y with Respect to X $y = - \csc ^ { - 1 } \left( \frac { 8 x + 9 } { 3 } \right)$