Question

question
find the derivative of (g(x)=3cos^{-1}(x)) at the point (x = \frac{1}{4}).
(enter an exact answer.)
provide your answer below:

Explanation:

Step1: Recall derivative formula

The derivative of $y = \cos^{-1}(u)$ is $y^\prime=-\frac{1}{\sqrt{1 - u^{2}}}\cdot u^\prime$. For $g(x)=3\cos^{-1}(x)$, using the constant - multiple rule of differentiation $(cf(x))^\prime = cf^\prime(x)$ where $c = 3$ and $f(x)=\cos^{-1}(x)$. So $g^\prime(x)=3\times(-\frac{1}{\sqrt{1 - x^{2}}})=-\frac{3}{\sqrt{1 - x^{2}}}$.

Step2: Evaluate at $x=\frac{1}{4}$

Substitute $x = \frac{1}{4}$ into $g^\prime(x)$. We get $g^\prime(\frac{1}{4})=-\frac{3}{\sqrt{1 - (\frac{1}{4})^{2}}}$. First, calculate $1-(\frac{1}{4})^{2}=1-\frac{1}{16}=\frac{15}{16}$. Then $\sqrt{1 - (\frac{1}{4})^{2}}=\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{4}$. So $g^\prime(\frac{1}{4})=-\frac{3}{\frac{\sqrt{15}}{4}}=-\frac{12}{\sqrt{15}}=-\frac{4\sqrt{15}}{5}$.

Answer:

$-\frac{4\sqrt{15}}{5}$