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)

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

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))

Simplify

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

= \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}) = e^{a x} a

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

= e^{a x} a 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)

\int_{1}^{3} x 2^{(x^{2})} d x

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

h \to \infty lim (h^{2} - 3 h)

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x)

x \to \frac{1}{2} lim (2 x - 1)

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)