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I have a senior 12th grade high school AP calculus One question about derivatives Picture included

I have a senior 12th grade high school AP calculus One question about derivatives-example-1
User Zadrozny
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1 Answer

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Given:

The equation of line is given as,


f(x)\text{ = cos}^(-1)(x)

Required:

Equation of tangent line at the point given as,


(\pi)/(2)

Step-by-step explanation:

The standard form of the equation of tangent line is given as,


y=mx+c

Slope of the required tangent line is calculated by taking the derivative of the given function.


Slope(m)\text{ = }(d)/(dx)\text{ \lparen cos}^(-1)(x))\text{ }

Therefore,


\begin{gathered} Slope(m)\text{ = }(-1)/(√(1-x^2)) \\ \end{gathered}

Slope at the given point is calculated as,


\begin{gathered} Slope(m)\text{ = }(-1)/(√(1-)((\pi)/(2))^2) \\ Slope(m)\text{ = }\frac{-1}{\sqrt{1-(\pi^2)/(4)}} \end{gathered}

On simplifying further,


Slope(m)\text{ = }(-2)/(√(2-\pi^2))

The tangent line passes through the point


((\pi)/(2),\text{ Undefined \rparen}

Therefore the equation of the tangent to the given line is,


y\text{ = \lbrack}(-2)/(√(2-\pi^2))]x\text{+ undefined}

Answer:

Thus the required equation of the tangent line is,


y=\frac{\text{-2}}{√(2-\pi^2)}x\text{ + undefined}

User Cardamo
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