What is the (e^tcos(t))^' ?

The (e^tcos(t))^' is e^tcos(t)-e^tsin(t)

What is the first (e^tcos(t))^' ?

The first (e^tcos(t))^' is e^tcos(t)-e^tsin(t)

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(e^{t} cos (t))^{^{'}}

e^{t} cos (t) - e^{t} sin (t)

Solution steps

(e^{t} cos (t))^{'}

Apply the Product Rule:

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

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

= (e^{t})^{^{'}} cos (t) + (cos (t))^{^{'}} e^{t}

(e^{t})^{'} = e^{t}

(cos (t))^{'} = - sin (t)

= e^{t} cos (t) + (- sin (t)) e^{t}

Simplify

= e^{t} cos (t) - e^{t} sin (t)

\frac{\partial}{\partial y} (y^{2} - x)

\frac{\partial}{\partial z} (x + y)

\int (\frac{6 x ^{2} - 3 x}{4 x ^{3}}) d x

2 (8^{x})

y^{^{''}} + 21 y = 0

What is the (e^tcos(t))^' ?
The (e^tcos(t))^' is e^tcos(t)-e^tsin(t)
What is the first (e^tcos(t))^' ?
The first (e^tcos(t))^' is e^tcos(t)-e^tsin(t)