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: $g(\frac{1}{4})=square$

Explanation:

Step1: Recall derivative formula

The derivative of $y = \cos^{- 1}(x)$ is $y^\prime=-\frac{1}{\sqrt{1 - x^{2}}}$. Using the constant - multiple rule, if $g(x)=3\cos^{-1}(x)$, then $g^\prime(x)=3\times(-\frac{1}{\sqrt{1 - x^{2}}})=-\frac{3}{\sqrt{1 - x^{2}}}$.

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 $1-(\frac{1}{4})^{2}=1-\frac{1}{16}=\frac{16 - 1}{16}=\frac{15}{16}$. 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:

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