What is the derivative of 2e^{5x}cos(6x) ?

The derivative of 2e^{5x}cos(6x) is 2(5e^{5x}cos(6x)-6e^{5x}sin(6x))

What is the first derivative of 2e^{5x}cos(6x) ?

The first derivative of 2e^{5x}cos(6x) is 2(5e^{5x}cos(6x)-6e^{5x}sin(6x))

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derivative of

2 e^{5 x} cos (6 x)

2 (5 e^{5 x} cos (6 x) - 6 e^{5 x} sin (6 x))

Solution steps

\frac{d}{d x} (2 e^{5 x} cos (6 x))

Take the constant out:

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

= 2 \frac{d}{d x} (e^{5 x} cos (6 x))

Apply the Product Rule:

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

f = e^{5 x}, g = cos (6 x)

= 2 (\frac{d}{d x} (e^{5 x}) cos (6 x) + \frac{d}{d x} (cos (6 x)) e^{5 x})

\frac{d}{d x} (e^{5 x}) = e^{5 x} \cdot 5

\frac{d}{d x} (cos (6 x)) = - sin (6 x) \cdot 6

= 2 (e^{5 x} \cdot 5 cos (6 x) + (- sin (6 x) \cdot 6) e^{5 x})

Simplify

= 2 (5 e^{5 x} cos (6 x) - 6 e^{5 x} sin (6 x))

\frac{1}{2} (t - 2)^{2}

\frac{\partial}{\partial x} (4 y + x e^{\frac{x}{y}})

y^{^{''}} - 4 y^{^{'}} + 5 y = 3 e^{2 t} sin (t)

- \frac{4 x}{x ^{2} + 1}

\int 7 sec^{2} (x) d x

What is the derivative of 2e^{5x}cos(6x) ?
The derivative of 2e^{5x}cos(6x) is 2(5e^{5x}cos(6x)-6e^{5x}sin(6x))
What is the first derivative of 2e^{5x}cos(6x) ?
The first derivative of 2e^{5x}cos(6x) is 2(5e^{5x}cos(6x)-6e^{5x}sin(6x))