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Popular >

derivative of e^{-ax}cos(bx)

Solution

\frac{d}{d x} (e^{- a x} cos (b x))

Solution

- a e^{- a x} cos (b x) - b e^{- a x} sin (b x)

Solution steps

\frac{d}{d x} (e^{- a x} cos (b x))

Apply the Product Rule:

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

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

= \frac{d}{d x} (e^{- a x}) cos (b x) + \frac{d}{d x} (cos (b x)) e^{- a x}

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

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

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

Simplify

= - a e^{- a x} cos (b x) - b e^{- a x} sin (b x)

Popular Examples

area y=|x|,y=x^2 area

y = ∣ x ∣, y = x^{2}

integral of (sin(θ)-1)/(cos^2(θ))

\int \frac{sin ( θ ) - 1}{cos ^{2} ( θ )} d θ

(\partial)/(\partial y)(e^{-xy})

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

(\partial)/(\partial x)(ye^{x/y}-z)

\frac{\partial}{\partial x} (y e^{\frac{x}{y}} - z)

derivative of (x^2/(5+8x))

\frac{d}{d x} (\frac{x ^{2}}{5 + 8 x})

Frequently Asked Questions (FAQ)

What is the derivative of e^{-ax}cos(bx) ?
The derivative of e^{-ax}cos(bx) is -ae^{-ax}cos(bx)-be^{-ax}sin(bx)
What is the first derivative of e^{-ax}cos(bx) ?
The first derivative of e^{-ax}cos(bx) is -ae^{-ax}cos(bx)-be^{-ax}sin(bx)