d/dx[√x] = d/dx[x^(1/2)] = (1/2)x^(-1/2)] = 1/(2√x)
d/dx[e√x] = e√x/(2√x)
d/dx[u/v] = [vu' - uv']/v^2
f(x) = e√x/x so u = e√x, v = x
f'(x) = [x*e√x/(2√x) - e√x]/x^2 = (√x - 2)[e√x]/2x^2
slope of the tangent, f'(4) = 0 because √x - 2 = 0
Regards - Ian
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
d/dx[√x] = d/dx[x^(1/2)] = (1/2)x^(-1/2)] = 1/(2√x)
d/dx[e√x] = e√x/(2√x)
d/dx[u/v] = [vu' - uv']/v^2
f(x) = e√x/x so u = e√x, v = x
f'(x) = [x*e√x/(2√x) - e√x]/x^2 = (√x - 2)[e√x]/2x^2
slope of the tangent, f'(4) = 0 because √x - 2 = 0
Regards - Ian