let $f(x)=6xsin^{-1}(x)$. a. differentiate $f(x)$. $f(x)=$ b. determine $f(0.5)$. $f(0.5)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u = 6x$ and $v=\sin^{-1}(x)$. The derivative of $u = 6x$ with respect to $x$ is $u'=6$, and the derivative of $v=\sin^{-1}(x)$ with respect to $x$ is $v'=\frac{1}{\sqrt{1 - x^{2}}}$. $f'(x)=6\sin^{-1}(x)+6x\cdot\frac{1}{\sqrt{1 - x^{2}}}$

Step2: Evaluate $f'(0.5)$

Substitute $x = 0.5$ into $f'(x)$. We know that $\sin^{-1}(0.5)=\frac{\pi}{6}$. $f'(0.5)=6\times\frac{\pi}{6}+6\times0.5\times\frac{1}{\sqrt{1-(0.5)^{2}}}$ $=\pi + 3\times\frac{1}{\sqrt{1 - 0.25}}=\pi+\frac{3}{\sqrt{0.75}}=\pi+\frac{3}{\frac{\sqrt{3}}{2}}=\pi + 2\sqrt{3}$

Answer:

a. $f'(x)=6\sin^{-1}(x)+\frac{6x}{\sqrt{1 - x^{2}}}$ b. $f'(0.5)=\pi + 2\sqrt{3}$