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

Question

3159 views around the world

You can reuse this answer
Creative Commons License

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−3exy = 4lnx + 2cosx - 3e^x?