Consider the function f(x)=tan(2x-pi/2).What is the derivative of f(x)=tan(2x-pi/2) What is the slope of the tangent to the curve f(x) at the point with x-coordinate 3pi/4 What is the slope of...

Consider the function f(x)=tan(2x-pi/2).

What is the derivative of f(x)=tan(2x-pi/2)

What is the slope of the tangent to the curve f(x) at the point with x-coordinate 3pi/4

What is the slope of tangent  of the tangent to the curve f(x) at the point with x-coordinate 3pi/4

 

Asked on by wang22

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The function f(x)=tan(2x - pi/2)

f(x)=tan(2x - pi/2)

f'(x) = `2*sec^2(2x - pi/2)`

At the point where the x-coordinate is `(3*pi)/4` the slope of the tangent to the curve f(x) is `f'(3pi/4) = 2*sec^2(2*3*pi/4 - pi/2)`

=> `2*sec^2(pi)`

=> 2

The derivative of `f(x)=tan(2x - pi/2)` is `f'(x) = 2*sec^2(2x - pi/2)` and at `x = 3*pi/4` the slope of the tangent is 2.

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should remember that tangent function is rational, hence `f(x) = (sin(2x - pi/2))/(cos(2x - pi/2))` .

You need to find derivative of `f(x), ` hence you need to use the quotient rule such that:

`f'(x) = ((sin(2x - pi/2))'*(cos(2x - pi/2)) - (sin(2x - pi/2))*(cos(2x - pi/2))')/(cos(2x - pi/2))^2`

`f'(x) = ((2cos(2x - pi/2))*(cos(2x - pi/2))+ 2(sin(2x - pi/2))*(sin(2x - pi/2))')/(cos(2x - pi/2))^2`

You need to factor out x such that:

`f'(x) = 2(cos^2(2x - pi/2) + sin^2(2x - pi/2))/(cos(2x - pi/2))^2`

You need to use the basic formula of trigonometry such that:

`sin^2 alpha + cos^2 alpha = 1`

`f'(x) = 2/(cos(2x - pi/2))^2`

You should remember that the slope of tangent line to curve f(x) is derivative at given point.

Hence, since you know the equation of derivative of function, you need to substitute `(3pi)/4 ` for x in equation of derivative such that:

`f'((3pi)/4)= 2/(cos(2(3pi/4) - pi/2))^2`

`f'((3pi)/4) = 2/(cos(3pi/2 - pi/2))^2`

`f'((3pi)/4) = 2/(cos(pi))^2`

`f'((3pi)/4) = 2/(-1)^2`

`f'((3pi)/4) = 2`

Hence, evaluating the derivative of the function f(x) and the slope of tangent line to the curve, at `x = 3pi/4` , f(x) yields `f'(x) = 2/(cos(2x - pi/2))^2`  and `f'((3pi)/4) = 2` .

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