How do you find the derivative of $y = .75 e^{x} + 2 cos x$ $y = .75e^x + 2 cos x$ ?

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Answer 1 · 2015-10-25T19:56:22

$\frac{d}{d x} (0, 75 e^{x} + 2 cos x) = 0, 75 e^{x} - 2 sin x$ $d/dx(0,75e^x+2cosx)=0,75e^x-2sinx$

By using laws of differentiation for exponential functions and trigonometric functions, we get

$\frac{d}{d x} (0, 75 e^{x} + 2 cos x) = 0, 75 e^{x} - 2 sin x$ $d/dx(0,75e^x+2cosx)=0,75e^x-2sinx$

How do you find the derivative of y = .75e^x + 2 cos xy=.75ex+2cosx y = .75e^x + 2 cos x?