What is the d/(dz)((2z+3i)(z-i)) ?

The d/(dz)((2z+3i)(z-i)) is 4z+i

What is the first d/(dz)((2z+3i)(z-i)) ?

The first d/(dz)((2z+3i)(z-i)) is 4z+i

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\frac{d}{d z} ((2 z + 3 i) (z - i))

4 z + i

Solution steps

\frac{d}{d z} ((2 z + 3 i) (z - i))

Simplify

(2 z + 3 i) (z - i) : (2 z^{2} + 3) + i z

= \frac{d}{d z} ((2 z^{2} + 3) + i z)

Apply the Sum/Difference Rule:

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

= \frac{d}{d z} (2 z^{2} + 3) + \frac{d}{d z} (i z)

\frac{d}{d z} (2 z^{2} + 3) = 4 z

\frac{d}{d z} (i z) = i

= 4 z + i

x \to 0 lim (\frac{3 x}{sin ( x )})

\int (x + 1) (x - x + 1) d x

\int t cos (t^{2}) d t

\int - \frac{1}{2 x ^{2}} d x

f (x) = (\frac{1}{2 x + 3})^{3}