$\text { Find the derivative of the function } f ( x ) = 10 \arcsin ( x - 9 ) \text {. }$
A)$f ^ { t } ( x ) = \frac { 10 } { \sqrt { ( x - 9 ) ^ { 2 } - 1 } }$
B) $f ^ { t } ( x ) = \frac { 10 } { ( x - 9 ) ^ { 2 } - 1 }$
C) $f ^ { t } ( x ) = \frac { 10 } { ( x - 9 ) ^ { 2 } + 1 }$
D) $f ^ { t } ( x ) = \frac { 10 } { \sqrt { 18 x - x ^ { 2 } - 80 } }$
E)$f ^ {t } ( x ) = \frac { 10 } { \sqrt { 1 + ( x - 9 ) ^ { 2 } } }$

Question

$\text { Find the derivative of the function } f ( x ) = 10 \arcsin ( x - 9 ) \text {. }$ 
A)$f ^ { t } ( x ) = \frac { 10 } { \sqrt { ( x - 9 ) ^ { 2 } - 1 } }$
B) $f ^ { t } ( x ) = \frac { 10 } { ( x - 9 ) ^ { 2 } - 1 }$
C) $f ^ { t } ( x ) = \frac { 10 } { ( x - 9 ) ^ { 2 } + 1 }$
D) $f ^ { t } ( x ) = \frac { 10 } { \sqrt { 18 x - x ^ { 2 } - 80 } }$
E)$f ^ {t } ( x ) = \frac { 10 } { \sqrt { 1 + ( x - 9 ) ^ { 2 } } }$

Quizplus · Accepted Answer

The derivative of $f(x) = 10 \arcsin(x - 9)$ is found using the chain rule. The derivative of $\arcsin(u)$ with respect to $u$ is $\frac{1}{\sqrt{1-u^2}}$, and the derivative of $u = x - 9$ with respect to $x$ is $1$. Thus, $f'(x) = \frac{10}{\sqrt{1 - (x - 9)^2}}$. Simplifying the expression under the square root gives $1 - (x^2 - 18x + 81) = 82 - 18x + x^2$, which rearranges to $18x - x^2 - 81$, matching the expression under the square root in option D.

Find the derivative of the function f(x)=10arcsin⁡(x−9). \text { Find the derivative of the function } f ( x ) = 10 \arcsin ( x - 9 ) \text {. } Find the derivative of the function f(x)=10arcsin(x−9).

$\text { Find the derivative of the function } f ( x ) = 10 \arcsin ( x - 9 ) \text {. }$