What is (\partial)/(\partial x)(e^{ax}cos(bx+y)) ?

The answer to (\partial)/(\partial x)(e^{ax}cos(bx+y)) is ae^{ax}cos(xb+y)-e^{ax}bsin(xb+y)

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(\partial)/(\partial x)(e^{ax}cos(bx+y))

\frac{\partial}{\partial x} (e^{a x} cos (b x + y))

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

Solution steps

\frac{\partial}{\partial x} (e^{a x} cos (b x + y))

Treat

a, b, y

as constants

Apply the Product Rule:

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

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

= \frac{\partial}{\partial x} (e^{a x}) cos (b x + y) + \frac{\partial}{\partial x} (cos (b x + y)) e^{a x}

\frac{\partial}{\partial x} (e^{a x}) = e^{a x} a

\frac{\partial}{\partial x} (cos (b x + y)) = - sin (b x + y) b

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

Simplify

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

\frac{d}{d x} (\frac{1 + x}{1 + y})

y = 49 - x^{2}, (- 7, 0)

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

\int ln (2 - x) d x

What is (\partial)/(\partial x)(e^{ax}cos(bx+y)) ?
The answer to (\partial)/(\partial x)(e^{ax}cos(bx+y)) is ae^{ax}cos(xb+y)-e^{ax}bsin(xb+y)