Answer:
[tex]\frac{1}{2\sqrt{5} }[/tex]
Step-by-step explanation:
Let, [tex]\text{sin}^{-1}(\frac{x}{6})[/tex] = y
sin(y) = [tex]\frac{x}{6}[/tex]
[tex]\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})[/tex]
[tex]\frac{d}{dx}\text{sin(y)}=\frac{1}{6}[/tex]
[tex]\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}[/tex] ---------(1)
[tex]\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}[/tex]
[tex]\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}[/tex]
cos(y) = [tex]\sqrt{1-\text{sin}^{2}(y) }[/tex]
= [tex]\sqrt{1-(\frac{x}{6})^2}[/tex]
= [tex]\sqrt{1-(\frac{x^2}{36})}[/tex]
Therefore, from equation (1),
[tex]\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}[/tex]
Or [tex]\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}[/tex]
At x = 4,
[tex]\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}[/tex]
[tex]\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}[/tex]
[tex]=\frac{1}{6\sqrt{\frac{36-16}{36}}}[/tex]
[tex]=\frac{1}{6\sqrt{\frac{20}{36} }}[/tex]
[tex]=\frac{1}{\sqrt{20}}[/tex]
[tex]=\frac{1}{2\sqrt{5}}[/tex]
