Find dy/dx and simplify the result, if possible

Question

a. 𝑦 = xe^𝑥 tan x b. 𝑦 = sin𝑥 / cos 𝑥 − 1 c. 𝑦 = 𝑥 + 1 / e^x

d. 𝑦 = 6x^2/3 - 3x^1/3 - x ^-1/3 e. 𝑦 = 𝑥^𝑒 − 𝑒^𝑥 − 𝜋^𝑒 + e

Answer 1

a. recursive chain rule: y = f(x)g(x)h(x) ->

y'(x) = f'(x)[g(x)h(x)] + f(x)[g'(x)h(x) + g(x)h'(x)] = f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)

where: f(x) = x, g(x) = e^x, h(x) = tan x, f'(x) = 1, g'(x) = e^x, h'(x) = sec² x

y'(x) = e^x tan x + xe^x tan x + xe^x sec² x = xe^x [tan x / x + tan x + sec² x] (one possible simplification)

b. quotient rule: y = f(x)/g(x) -> y'(x) = [f'(x)g(x) - f(x)g'(x)] / [g(x)]²

where: f(x) = sin x, g(x) = cos x - 1, f'(x) = cos x, g'(x) = -sin x

y'(x) = cos² x - cos x + sin² x / [cos x - 1]²

use: cos² x + sin² x = 1 to simplify numerator

y'(x) = [1 - cos x] / [cos x - 1]² = -1 / [cos x - 1] = 1 / [1 - cos x]

c. quotient rule: y = f(x)/g(x) -> y'(x) = [f'(x)g(x) - f(x)g'(x)] / [g(x)]²

where: f(x) = x + 1, g(x) = e^x, f'(x) = 1, g'(x) = e^x

y'(x) = e^x - (x + 1)e^x / e^2x = -xe^x / e^2x = -x / e^x

d. power rule: y(x) = xⁿ -> y'(x) = nx^n-1

y'(x) = 4x^-1/3 - x^-2/3 + (1/3)x^-4/3

e. use power rule, exponential rule (d/dx e^x = e^x) , and constant rule (d/dx constant = 0) respectively:

y'(x) = ex^e-1 - e^x