What is the d/(da)(4sin(3a)cos(3a)) ?

The d/(da)(4sin(3a)cos(3a)) is 12cos(6a)

What is the first d/(da)(4sin(3a)cos(3a)) ?

The first d/(da)(4sin(3a)cos(3a)) is 12cos(6a)

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\frac{d}{d a} (4 sin (3 a) cos (3 a))

12 cos (6 a)

Solution steps

\frac{d}{d a} (4 sin (3 a) cos (3 a))

Simplify

4 sin (3 a) cos (3 a) : 2 sin (6 a)

= \frac{d}{d a} (2 sin (6 a))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d a} (sin (6 a))

Apply the chain rule:

cos (6 a) \frac{d}{d a} (6 a)

= cos (6 a) \frac{d}{d a} (6 a)

\frac{d}{d a} (6 a) = 6

= 2 cos (6 a) \cdot 6

Simplify

= 12 cos (6 a)

\frac{d}{d x} (ln (x) d x)

x \to 0 + lim (\frac{8}{x ^{2} - x})

(8.4) (4.1)

x \to - 1 - lim (f (x))

\frac{d}{d x} (\frac{a ^{2} + x ^{2}}{x})

What is the d/(da)(4sin(3a)cos(3a)) ?
The d/(da)(4sin(3a)cos(3a)) is 12cos(6a)
What is the first d/(da)(4sin(3a)cos(3a)) ?
The first d/(da)(4sin(3a)cos(3a)) is 12cos(6a)