What is the derivative of axsin(4x+bxcos(4x)) ?

The derivative of axsin(4x+bxcos(4x)) is a(sin(4x)+4xcos(4x))+b(cos(4x)-4xsin(4x))

What is the first derivative of axsin(4x+bxcos(4x)) ?

The first derivative of axsin(4x+bxcos(4x)) is a(sin(4x)+4xcos(4x))+b(cos(4x)-4xsin(4x))

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\frac{d}{d x} (a x sin (4 x) + b x cos (4 x))

a (sin (4 x) + 4 x cos (4 x)) + b (cos (4 x) - 4 x sin (4 x))

Solution steps

\frac{d}{d x} (a x sin (4 x) + b x cos (4 x))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (a x sin (4 x)) + \frac{d}{d x} (b x cos (4 x))

\frac{d}{d x} (a x sin (4 x)) = a (sin (4 x) + 4 x cos (4 x))

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

= a (sin (4 x) + 4 x cos (4 x)) + b (cos (4 x) - 4 x sin (4 x))

f^{^{'}} (t) - f (t) = 2 t

\int e^{2 x + 1} d x

x \to 3 + lim (\frac{- x}{x - 3})

x \to 8 lim (\frac{3}{( x - 8 ) ^{2}})

\int 13 tan^{4} (x) sec^{6} (x) d x

What is the derivative of axsin(4x+bxcos(4x)) ?
The derivative of axsin(4x+bxcos(4x)) is a(sin(4x)+4xcos(4x))+b(cos(4x)-4xsin(4x))
What is the first derivative of axsin(4x+bxcos(4x)) ?
The first derivative of axsin(4x+bxcos(4x)) is a(sin(4x)+4xcos(4x))+b(cos(4x)-4xsin(4x))