Find the derivative of y with respect to x.
-$$y = \sin ^ { - 1 } \left( \frac { 1 } { x ^ { 4 } } \right)$$
A) $\frac { - 4 x ^ { 4 } } { \sqrt { 1 - x ^ { 8 } } }$
B) $\frac { - 4 } { x \sqrt { x ^ { 8 } - 1 } }$
C) $\frac { - 4 } { 1 + x ^ { 8 } }$
D) $\frac { - 4 } { x \sqrt { 1 - x ^ { 8 } } }$

Question

Quizplus · Accepted Answer

To find the derivative of $y = \sin^{-1}\left(\frac{1}{x^4}\right)$ with respect to $x$, we use the chain rule. The derivative of $\sin^{-1}(u)$ with respect to $u$ is $\frac{1}{\sqrt{1-u^2}}$, and the derivative of $\frac{1}{x^4}$ with respect to $x$ is $-4x^{-5}$. Applying the chain rule, we get $\frac{d}{dx}y = \frac{1}{\sqrt{1-\left(\frac{1}{x^4}\right)^2}} \cdot (-4x^{-5}) = \frac{-4}{x^5\sqrt{1-\frac{1}{x^8}}} = \frac{-4}{x\sqrt{x^8-1}}$.

Find the Derivative of Y with Respect to X y=sin⁡−1(1x4)y = \sin ^ { - 1 } \left( \frac { 1 } { x ^ { 4 } } \right)y=sin−1(x41​)

Find the Derivative of Y with Respect to X $y = \sin ^ { - 1 } \left( \frac { 1 } { x ^ { 4 } } \right)$