How do you differentiate $y = 4lnx + 2cosx - 3e^x$ ?

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Answer 1 · 2015-12-19T13:48:20

$f'(x) = 4/x -2sin(x) - 3e^x$

You simply differentiate term by term and then sum the derivatives.

$(4ln(x))' = 4*1/x; (2cos(x))' = 2*(-sin(x); (-3e^x)' = -3*e^x$ .

How do you differentiate y = 4lnx + 2cosx - 3e^xy = 4lnx + 2cosx - 3e^x?