What is the (e^{-x}sin(x))^' ?

The (e^{-x}sin(x))^' is -e^{-x}sin(x)+e^{-x}cos(x)

What is the first (e^{-x}sin(x))^' ?

The first (e^{-x}sin(x))^' is -e^{-x}sin(x)+e^{-x}cos(x)

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(e^{- x} sin (x))^{^{'}}

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

Solution steps

(e^{- x} sin (x))^{'}

Apply the Product Rule:

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

f = e^{- x}, g = sin (x)

= (e^{- x})^{^{'}} sin (x) + (sin (x))^{^{'}} e^{- x}

(e^{- x})^{'} = - e^{- x}

(sin (x))^{'} = cos (x)

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

Simplify

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

\int 4 x cos (x) d x

\int_{- 2}^{3} \frac{3}{x ^{4}} d x

\frac{\partial}{\partial x} (v)

n = 0 \sum \infty (- 1)^{n} x^{3 n}

(x, y) \to (1, 1) lim (x + y)

What is the (e^{-x}sin(x))^' ?
The (e^{-x}sin(x))^' is -e^{-x}sin(x)+e^{-x}cos(x)
What is the first (e^{-x}sin(x))^' ?
The first (e^{-x}sin(x))^' is -e^{-x}sin(x)+e^{-x}cos(x)