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

Answer 1

Samantha Adkison

#d/dx(0,75e^x+2cosx)=0,75e^x-2sinx#

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

#d/dx(0,75e^x+2cosx)=0,75e^x-2sinx#

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Answer 2

Emily Ammerman

To find the derivative of ( y = 0.75e^x + 2\cos(x) ), you can apply the derivative rules as follows:

[ \frac{dy}{dx} = 0.75\frac{d}{dx}(e^x) + 2\frac{d}{dx}(\cos(x)) ]

[ = 0.75e^x - 2\sin(x) ]

So, the derivative of ( y ) with respect to ( x ) is ( \frac{dy}{dx} = 0.75e^x - 2\sin(x) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.