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

The (e^tsin(t))^' is e^tsin(t)+cos(t)e^t

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

The first (e^tsin(t))^' is e^tsin(t)+cos(t)e^t

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

\frac{\partial}{\partial v} (u (v) (v)^{2 v} + e^{3})

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 2, y^{^{'}} (0) = 2

\frac{d}{d x} (x^{3} - 2 x^{2} + 1)

e^{(- 2 s)}

\frac{d y}{d x} = 4 x e^{6 y}

What is the (e^tsin(t))^' ?
The (e^tsin(t))^' is e^tsin(t)+cos(t)e^t
What is the first (e^tsin(t))^' ?
The first (e^tsin(t))^' is e^tsin(t)+cos(t)e^t