calculus help!!!!?

Question

Given f(x)=x^2tanx, write an equation of the line tangen to the graph at x=pi/4

Answer 1

We need this: d/dx(tan x) = (sec x)^2. So:

y = f(x) = x^2 tan x
dy/dx = x^2 (sec x)^2 + 2x tan x

For x = pi/4:

y = (pi^2 / 16) (1) = pi^2 / 16

dy/dx = (pi^2 / 16) (sqrt 2)^2 + 2(pi/4)(1)
= pi^2 / 8 + pi / 2

Using point-slope:

y - pi^2 / 16 = (pi^2 / 8 + pi / 2) (x - pi/4)

y = (pi^2 / 8 + pi / 2)x - pi^3 / 32 - pi^2 / 8 + pi^2 / 16

y = (pi^2 / 8 + pi / 2)x - pi^3 / 32 - pi^2 / 16 (Answer)

You can clean this up a bit by factoring out pi / 32:

y = (pi/32) [(4 pi + 16)x - pi^2 - 2 pi] (Alternative answer)

You can also plug in the numbers to get:

y = 2.8045 x - 1.5858 (Another alternative answer)

As a check, plug in x = pi/4 = 0.7854. Then y = 0.6169, and that matches the value of your original function f(x) at that point. So the answer is right.

Answer 2

although i will not give you the answer i will tell you how to get it your self.

Use the product rule for f(x) (f'(x)=f'g+g'f) then plug in Pi/4 for x then solve that gives you the slope of the tangent line. then use y=mx+b to get the equation of the tangent line of f(x)=x^2tanx. hope this helps