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If f (x) = sin−1x, then f′(√3/2)= ?

Sagot :

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.

Cricetus

The value of "[tex]f'(\sqrt{\frac{3}{2} } )[/tex]" will be "2".

Given that:

  • [tex]f(x) = sin^{-1} x[/tex]

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

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