What is the derivative of y=8e^xcos(x) ?

The derivative of y=8e^xcos(x) is 8(e^xcos(x)-e^xsin(x))

What is the first derivative of y=8e^xcos(x) ?

The first derivative of y=8e^xcos(x) is 8(e^xcos(x)-e^xsin(x))

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derivative of

y = 8 e^{x} cos (x)

8 (e^{x} cos (x) - e^{x} sin (x))

Solution steps

\frac{d}{d x} (8 e^{x} cos (x))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d x} (e^{x} cos (x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x}, g = cos (x)

= 8 (\frac{d}{d x} (e^{x}) cos (x) + \frac{d}{d x} (cos (x)) e^{x})

\frac{d}{d x} (e^{x}) = e^{x}

\frac{d}{d x} (cos (x)) = - sin (x)

= 8 (e^{x} cos (x) + (- sin (x)) e^{x})

Simplify

= 8 (e^{x} cos (x) - e^{x} sin (x))

\int \frac{( x + 1 )}{x} d x

\int \frac{1}{e ^{2 x} + 1} d x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (x), y (0) = 1, y^{^{'}} (0) = - 1

y^{^{'}} = x + y, y (0) = 7

\frac{d}{d x} (\frac{4 x ^{2} + 2 x + 8}{x})

What is the derivative of y=8e^xcos(x) ?
The derivative of y=8e^xcos(x) is 8(e^xcos(x)-e^xsin(x))
What is the first derivative of y=8e^xcos(x) ?
The first derivative of y=8e^xcos(x) is 8(e^xcos(x)-e^xsin(x))