What is the derivative of Asin(2x+Bcos(2x)) ?

The derivative of Asin(2x+Bcos(2x)) is Acos(2x)*2-2Bsin(2x)

What is the first derivative of Asin(2x+Bcos(2x)) ?

The first derivative of Asin(2x+Bcos(2x)) is Acos(2x)*2-2Bsin(2x)

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\frac{d}{d x} (A sin (2 x) + B cos (2 x))

A cos (2 x) \cdot 2 - 2 B sin (2 x)

Solution steps

\frac{d}{d x} (A sin (2 x) + B cos (2 x))

Treat

A, B

as constants

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (A sin (2 x)) + \frac{d}{d x} (B cos (2 x))

\frac{d}{d x} (A sin (2 x)) = A cos (2 x) \cdot 2

\frac{d}{d x} (B cos (2 x)) = - 2 B sin (2 x)

= A cos (2 x) \cdot 2 - 2 B sin (2 x)

y = 12 cos (x)

\frac{d}{d t} (k t)

\int_{- 1}^{3} 6 x^{2} d x

e^{- x^{2}}

\frac{e ^{x}}{6 x ^{2} + 7}

What is the derivative of Asin(2x+Bcos(2x)) ?
The derivative of Asin(2x+Bcos(2x)) is Acos(2x)*2-2Bsin(2x)
What is the first derivative of Asin(2x+Bcos(2x)) ?
The first derivative of Asin(2x+Bcos(2x)) is Acos(2x)*2-2Bsin(2x)