Find the derivative of y with respect to x.
-$$y = 2 x ^ { 4 } \sin ^ { - 1 } x$$
A) $\frac { 2 x ^ { 4 } } { \sqrt { 1 - x ^ { 2 } } } + 8 x ^ { 3 } \sin ^ { - 1 } x$
B) $\frac { 2 x ^ { 4 } } { \sqrt { 1 - x ^ { 2 } } }$
C) $\frac { 2 x ^ { 4 } } { 1 + x ^ { 2 } } + 8 x ^ { 3 } \sin ^ { - 1 } x$
D) $\frac { 1 } { \sqrt { 1 - x ^ { 2 } } } + 8 x ^ { 3 }$

Question

Quizplus · Accepted Answer

Use the product rule and chain rule. Let 
$$f(x)=2x^4$$
$$g(x)=\sin^{-1}x$$
Then,
$$f'(x)=8x^3$$
$$g'(x)=\frac{1}{\sqrt{1-x^2}}$$
By the product rule,
$$y'=f(x)g'(x)+f'(x)g(x)=2x^4\frac{1}{\sqrt{1-x^2}}+8x^3\sin^{-1}x$$
Simplify to get the answer A.

Find the Derivative of Y with Respect to X y=2x4sin⁡−1xy = 2 x ^ { 4 } \sin ^ { - 1 } xy=2x4sin−1x

Find the Derivative of Y with Respect to X $y = 2 x ^ { 4 } \sin ^ { - 1 } x$