What is the d/(dy)(-2e^x+4xcos(y)) ?

The d/(dy)(-2e^x+4xcos(y)) is -4xsin(y)

What is the first d/(dy)(-2e^x+4xcos(y)) ?

The first d/(dy)(-2e^x+4xcos(y)) is -4xsin(y)

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\frac{d}{d y} (- 2 e^{x} + 4 x cos (y))

- 4 x sin (y)

Solution steps

\frac{d}{d y} (- 2 e^{x} + 4 x cos (y))

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d y} (2 e^{x}) + \frac{d}{d y} (4 x cos (y))

\frac{d}{d y} (2 e^{x}) = 0

\frac{d}{d y} (4 x cos (y)) = - 4 x sin (y)

= - 0 - 4 x sin (y)

Simplify

= - 4 x sin (y)

\int 2 x - 8 y d x

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\int tan^{3} (11 π x) d x

\frac{d}{d x} (\frac{1}{4} sin (2 x))

What is the d/(dy)(-2e^x+4xcos(y)) ?
The d/(dy)(-2e^x+4xcos(y)) is -4xsin(y)
What is the first d/(dy)(-2e^x+4xcos(y)) ?
The first d/(dy)(-2e^x+4xcos(y)) is -4xsin(y)