Answer:
[tex]f'(\dfrac{\sqrt3}{2})[/tex] = 2
Explanation:
Given that,
[tex]f(x)=\sin^{-1} x[/tex]
We need to find the value of [tex]f'(\dfrac{\sqrt3}{2})[/tex]
First, we take f'(x). We know that,
[tex]\dfrac{d}{dx}(\sin^{-1} x)=\dfrac{1}{\sqrt{1-x^2} }\\\\f'(x)=\dfrac{1}{\sqrt{1-x^2} }\\\\f'(\dfrac{\sqrt3}{2})=\dfrac{1}{\sqrt{1-(\dfrac{\sqrt3}{2})^2} }\\\\=\dfrac{1}{\sqrt{1-(\dfrac{3}{4})} }\\\\=\dfrac{1}{\sqrt{\dfrac{1}{4}}}\\\\=\dfrac{1}{\dfrac{1}{2}}\\\\=2[/tex]
So, the value of [tex]f'(\dfrac{\sqrt3}{2})[/tex] is 2.
Given that:
We know that,
→ [tex]\frac{d}{dx} (sin^{-1} x) = \frac{1}{\sqrt{1-x^2} }[/tex]
Now,
→ [tex]f'(x) = \frac{1}{\sqrt{1-x^2} }[/tex]
then,
→ [tex]f'(\frac{\sqrt{3} }{2} ) = \frac{1}{1-(\frac{\sqrt{3} }{2} )^2}[/tex]
By opening the square, we get
[tex]= \frac{1}{\sqrt{1-(\frac{2}{4} )} }[/tex]
[tex]= \frac{1}{\sqrt{\frac{1}{4} } }[/tex]
[tex]= \frac{1}{\frac{1}{2} }[/tex]
[tex]= 1\times \frac{2}{1}[/tex]
[tex]= 2[/tex]
Thus the above answer is right.
Learn more about function f(x) here:
https://brainly.com/question/17032105
