Question

How do you find the derivative of `2sinx-tanx`?

Answer 1

`(d(2sinx -tanx))/(dx) = 2cos x -1/(cos^2x)`

Explanation:

As the derivative of a function is a linear operator, we know that:

`(d(f+g))/(dx) = (df)/(dx)+(dg)/(dx)`

and

`(d(lambda f))/(dx) = lambda(df)/(dx)`.

We have therefore:

`(d(2sinx -tanx))/(dx) = 2(d(sinx))/(dx)-(d(tanx))/(dx) = 2cos x -1/(cos^2x)`