Given:
The equation of line is given as,
[tex]f(x)\text{ = cos}^{-1}(x)[/tex]
Required:
Equation of tangent line at the point given as,
[tex]\frac{\pi}{2}[/tex]
Explanation:
The standard form of the equation of tangent line is given as,
[tex]y=mx+c[/tex]
Slope of the required tangent line is calculated by taking the derivative of the given function.
[tex]Slope(m)\text{ = }\frac{d}{dx}\text{ \lparen cos}^{-1}(x))\text{ }[/tex]
Therefore,
[tex]\begin{gathered} Slope(m)\text{ = }\frac{-1}{\sqrt{1-x^2}} \\ \end{gathered}[/tex]
Slope at the given point is calculated as,
[tex]\begin{gathered} Slope(m)\text{ = }\frac{-1}{\sqrt{1-}(\frac{\pi}{2})^2} \\ Slope(m)\text{ = }\frac{-1}{\sqrt{1-\frac{\pi^2}{4}}} \end{gathered}[/tex]
On simplifying further,
[tex]Slope(m)\text{ = }\frac{-2}{\sqrt{2-\pi^2}}[/tex]
The tangent line passes through the point
[tex](\frac{\pi}{2},\text{ Undefined \rparen}[/tex]
Therefore the equation of the tangent to the given line is,
[tex]y\text{ = \lbrack}\frac{-2}{\sqrt{2-\pi^2}}]x\text{+ undefined}[/tex]
Answer:
Thus the required equation of the tangent line is,
[tex]y=\frac{\text{-2}}{\sqrt{2-\pi^2}}x\text{ + undefined}[/tex]
