given f(x)=2sec^2(x) , write the equation of the line tangent to y=f(x) when x=7pi/4

Question

HAS TO BE IN y-y1=m(x-x1) FORM

Answer 1

So first thing you need to do is plug in 7pi/4 into 2sec^2(x) to get your point for slope point

2sec^2(7pi/4) = 2*(√2)^2 = 4

point: (7pi/4,4)

Now, we need our slope! We first need to take the derivative of 2sec^2(x) which can be done using chain rule.

Let's make u = sec(x) and take the derivative of that to get u'=sec(x)tan(x).

f(x) = 2sec^2(x)

= 2u^2 We rewrite the expression in terms of u

Now we apply chain rule

f'(x) = 4u * u'

= 4(sec(x))(sec(x))(tan(x)) and we just plug in our u and u'

= 4sec^2(x)tan(x)

Now we substitute in 7pi/4 to get our slope

f'(7pi/4) = 4sec(7pi/4)^2*tan(7pi/4) = 4*(sqrt(2)^2)*-1 = - 8

So our final tangent line using the point and slope we got is y-4 = -8(x-7pi/4)

Answer 2

dy/dx=sec²x tan x and at x=7π/4 dy/dx=-2

The function value at 7π/4 is 4.

Therefore, the equation of the tangent line is -2=(y-4)/[x-(7π/4)].

I will leave it to you to get it into the form required.