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

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Answer 1 · 2016-12-12T09:24:30

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

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:

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

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