What is the d/(db)(sin^4(2b)-cos^4(2b)) ?

The d/(db)(sin^4(2b)-cos^4(2b)) is 8cos(2b)sin(2b)

What is the first d/(db)(sin^4(2b)-cos^4(2b)) ?

The first d/(db)(sin^4(2b)-cos^4(2b)) is 8cos(2b)sin(2b)

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\frac{d}{d b} (sin^{4} (2 b) - cos^{4} (2 b))

8 cos (2 b) sin (2 b)

Solution steps

\frac{d}{d b} (sin^{4} (2 b) - cos^{4} (2 b))

Apply the Sum/Difference Rule:

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

= \frac{d}{d b} (sin^{4} (2 b)) - \frac{d}{d b} (cos^{4} (2 b))

\frac{d}{d b} (sin^{4} (2 b)) = 8 sin^{3} (2 b) cos (2 b)

\frac{d}{d b} (cos^{4} (2 b)) = - 8 cos^{3} (2 b) sin (2 b)

= 8 sin^{3} (2 b) cos (2 b) - (- 8 cos^{3} (2 b) sin (2 b))

Simplify

8 sin^{3} (2 b) cos (2 b) - (- 8 cos^{3} (2 b) sin (2 b)) : 8 cos (2 b) sin (2 b)

= 8 cos (2 b) sin (2 b)

2 x^{2} y - x^{3} y^{^{'}} = y^{3}

\int \frac{x ^{2}}{1 + x ^{6}} d x

\int \frac{x ^{2}}{a} e^{- \frac{x}{a}} d x

\frac{\partial}{\partial x} (e^{(t - x)})

\frac{d}{d x} (4 x^{2} - 5)

What is the d/(db)(sin^4(2b)-cos^4(2b)) ?
The d/(db)(sin^4(2b)-cos^4(2b)) is 8cos(2b)sin(2b)
What is the first d/(db)(sin^4(2b)-cos^4(2b)) ?
The first d/(db)(sin^4(2b)-cos^4(2b)) is 8cos(2b)sin(2b)