Question

determine dx/dy:
1)y=(sec x+3 ln tanx)^1/4
2)y=lnsqrte^3x(x^2-1)

Answer 1

@mkrugel Are you familiar with the chain rule?

Answer 2

For question one,
y=(sec x+3 ln tanx)^1/4

Let u = sec x+3 ln tanx
So, y = u^(1/4)

\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]
Find dy/du and du/dx respectively. Can you do it now?