How to find dy/dx of given functions

Question

How do I find dy/dx of the following functions:

i) y = (e^x + 1/x)ln(x² - e^2x)

ii) y = (∛(x²) + √(x))/x

For both do not simplify the answer

Answer 1

Hi Chance,

For i) y = (e^x + 1/x)ln(x² - e^2x).

Use the product rule: dy/dx = d(uv)/dx = u(dv/dx) + v(du/dx).

Let u = (e^x + 1/x) and v = ln(x² - e^2x).

dy/dx = u(d[ln((x² - e^2x)]/dx) + v(d[(e^x + 1/x)]/dx).

Let's evaluate the second term first as this is simpler.

u(d[ln((x² - e^2x)]/dx) + v(e^x - (1/x²)).

To deal with the first term, we need to use the chain rule: (dy/dx)=(dy/dz)(dz/dx).

Let z = (x² - e^2x).

d[ln((x² - e^2x)]/dx = (dy/dz)(dz/dx) = (d[ln(z)]/dz)(d[x² - e^2x]/dx) =

(1/z)*(2x - 2e^2x).

Plug this back into: u(d[ln((x² - e^2x)]/dx) + v(e^x - (1/x²)). --> u[(1/z)*(2x - 2e^2x)] + v(e^x - (1/x²)).

Replace z with (x² - e^2x).

u[(1/(x² - e^2x))*(2x - 2e^2x)] + v(e^x - (1/x²)).

Replace u = (e^x + 1/x) and v = ln(x² - e^2x).

dy/dx = [(e^x + 1/x)*(2x - 2e^2x)/(x² - e^2x)] + [(ln(x² - e^2x))*(e^x - (1/x²))].

Answer 2

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

dy/dx = y'= (2/3)x^(-1/3) - (1/2)x^(-3/2)