$\text { Find the derivative of the function } f ( x ) = 9 \arcsin \left( 6 x ^ { 2 } + 19 x - 9 \right)$
A)$f ^ { t } ( x ) = \frac { 9 } { 1 + \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } }$
B)$f ^ { t } ( x ) = \frac { 9 ( 12 x + 19 ) } { 1 + \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } }$
C)$f ^ { t } ( x ) = \frac { 9 ( 12 x + 19 ) } { \sqrt { 1 - \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } } }$
D)$f ^ { t} ( x ) = \frac { 9 } { \sqrt { 1 - \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } } }$
E)$f ^ { t } ( x ) = \frac { 9 ( 12 x + 19 ) } { \sqrt { 1 + \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } } }$

Question

$\text { Find the derivative of the function } f ( x ) = 9 \arcsin \left( 6 x ^ { 2 } + 19 x - 9 \right)$ 
A)$f ^ { t } ( x ) = \frac { 9 } { 1 + \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } }$
B)$f ^ { t } ( x ) = \frac { 9 ( 12 x + 19 ) } { 1 + \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } }$
C)$f ^ { t } ( x ) = \frac { 9 ( 12 x + 19 ) } { \sqrt { 1 - \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } } }$
D)$f ^ { t} ( x ) = \frac { 9 } { \sqrt { 1 - \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } } }$
E)$f ^ { t } ( x ) = \frac { 9 ( 12 x + 19 ) } { \sqrt { 1 + \left( 6 x ^ { 2 } + 19 x - 9 \right) ^ { 2 } } }$

Quizplus · Accepted Answer

The derivative of $f(x) = 9\arcsin(g(x))$, where $g(x) = 6x^2 + 19x - 9$, is given by the chain rule as $f'(x) = \frac{9g'(x)}{\sqrt{1 - g(x)^2}}$. Here, $g'(x) = 12x + 19$, so the correct derivative is $f'(x) = \frac{9(12x + 19)}{\sqrt{1 - (6x^2 + 19x - 9)^2}}$.

Find the derivative of the function f(x)=9arcsin⁡(6x2+19x−9)\text { Find the derivative of the function } f ( x ) = 9 \arcsin \left( 6 x ^ { 2 } + 19 x - 9 \right) Find the derivative of the function f(x)=9arcsin(6x2+19x−9)

$\text { Find the derivative of the function } f ( x ) = 9 \arcsin \left( 6 x ^ { 2 } + 19 x - 9 \right)$