# `h(t) = sec(t)/t, (pi,-1/pi)` Evaluate the derivative of the function at the given point. Use a graphing utility to verify your result.

### Textbook Question

Chapter 2, 2.3 - Problem 61 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the equation of the tangent line to the curve `f(t) =(sec t)/t` , at the point `(pi, -1/(pi)), ` using the following formula, such that:

`f(t) - f(pi) = f'(pi)(t - pi)`

Notice that `f(pi)=1/(pi).`

You need to evaluate f'(t), using the quotient rule, and then `f'(pi):`

`f'(t) = ((sec t)'*t- (sec t)*(t)')/(t^2)`

`f'(t) = (t*sec t*tan t - sec t)/(t^2)`

`f'(pi)= (pi*1/(cos pi)* tan pi - 1/(cos pi))/(pi^2)`

`cos pi = -1 and tan pi = 0`

`f'(pi)= (1)/(pi^2)`

You need to replace the values into the equation of tangent line:

`f(t) - 1/(pi) = (1)/(pi^2)*(t - pi)`

`f(t) = 1/(pi) + t/(pi^2) - 1/(pi)`

reducing like terms yields:

`f(t) = t/(pi^2) `

Hence, evaluating the equation of the tangent line to te given curve , at the given point, yields `f(t) = t/(pi^2) ` .