Find the equation of the tangent line to f(x)=x sec x at (pi/4, pi •sqrt of 2 all over 4).
I got to here, but did not know what to do next:
slope: (pi/4)(sqrt of 2 over 2) + (sqrt of 2 over 2) everything over 1/2=.......
y = x secx , point is (π/4 , π√2/4)
dy/dx = x secx tanx + secx
= secx(x tanx + 1)
so at the given point:
slope = √2( (π/4)(1) + 1) = π√2/4 + √2
so using y = mx + b
π√2/4 = (π√2/4 + √2)(π/4) + b
solve this mess for b and you are done
looks good to me
http://www.wolframalpha.com/input/?i=plot+y+%3D+xsecx%2C+y+%3D+%28%CF%80%E2%88%9A2%2F4+%2B+%E2%88%9A2%29x+%2B+%CF%80%E2%88%9A2%2F4+-+%28%CF%80%E2%88%9A2%2F4+%2B+%E2%88%9A2%29%28%CF%80%2F4%29
To find the equation of the tangent line to f(x) = x sec x at the point (pi/4, pi • sqrt(2)/4), we need to determine the slope of the tangent line and the point of tangency.
To find the slope of the tangent line, we can use the derivative of the function f(x). The derivative of f(x) can be found using the product rule and the chain rule. The derivative of x sec x with respect to x is given by:
f'(x) = (1)sec x + x(sec x)(tan x)
= sec x + x(sec x)(tan x)
Now evaluate f'(x) at x = pi/4:
f'(pi/4) = sec(pi/4) + pi/4(sec(pi/4))(tan(pi/4))
We know that sec(pi/4) = sqrt(2), and tan(pi/4) = 1, so we can substitute these values into the equation:
f'(pi/4) = sqrt(2) + pi/4(sqrt(2))(1)
= sqrt(2) + pi•sqrt(2)/4
This gives us the slope of the tangent line at x = pi/4.
Now that we have the slope, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is:
y - y1 = m(x - x1)
Using the slope we found (m = sqrt(2) + pi•sqrt(2)/4) and the given point (x1 = pi/4, y1 = pi•sqrt(2)/4):
y - pi•sqrt(2)/4 = (sqrt(2) + pi•sqrt(2)/4)(x - pi/4)
Now simplify the equation:
y - pi•sqrt(2)/4 = sqrt(2)x + (pi•sqrt(2)/4)x - sqrt(2)(pi/4) - (pi/4)sqrt(2)
Combine like terms:
y = sqrt(2)x + (pi•sqrt(2)/4)x + pi•sqrt(2)/4 - sqrt(2)(pi/4) - (pi/4)sqrt(2)
Simplify further:
y = (sqrt(2) + pi•sqrt(2)/4)x + pi•sqrt(2)/4 - pi•sqrt(2)/4 - pi•sqrt(2)/4
y = (sqrt(2) + pi•sqrt(2)/4)x
Therefore, the equation of the tangent line to f(x) = x sec x at the point (pi/4, pi •sqrt(2)/4) is y = (sqrt(2) + pi•sqrt(2)/4)x.