What is the d/(dz)(2^zlog_{2}(z)) ?

The d/(dz)(2^zlog_{2}(z)) is 2^zln(z)+(2^z)/(ln(2)z)

What is the first d/(dz)(2^zlog_{2}(z)) ?

The first d/(dz)(2^zlog_{2}(z)) is 2^zln(z)+(2^z)/(ln(2)z)

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\frac{d}{d z} (2^{z} lo g_{2} (z))

2^{z} ln (z) + \frac{2 ^{z}}{ln ( 2 ) z}

Solution steps

\frac{d}{d z} (2^{z} lo g_{2} (z))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 2^{z}, g = lo g_{2} (z)

= \frac{d}{d z} (2^{z}) lo g_{2} (z) + \frac{d}{d z} (lo g_{2} (z)) 2^{z}

\frac{d}{d z} (2^{z}) = 2^{z} ln (2)

\frac{d}{d z} (lo g_{2} (z)) = \frac{1}{z ln ( 2 )}

= 2^{z} ln (2) lo g_{2} (z) + \frac{1}{z ln ( 2 )} \cdot 2^{z}

Simplify

2^{z} ln (2) lo g_{2} (z) + \frac{1}{z ln ( 2 )} \cdot 2^{z} : 2^{z} ln (z) + \frac{2 ^{z}}{ln ( 2 ) z}

= 2^{z} ln (z) + \frac{2 ^{z}}{ln ( 2 ) z}

\frac{d}{d x} (\frac{x ^{2} - 3}{x ^{3}})

3 x^{^{'}} + 6 x = 4

\frac{r d θ}{d r} = θ^{2} + 1

f (x) = 5 - x^{2}, at x = 1

x^{2} - 7

What is the d/(dz)(2^zlog_{2}(z)) ?
The d/(dz)(2^zlog_{2}(z)) is 2^zln(z)+(2^z)/(ln(2)z)
What is the first d/(dz)(2^zlog_{2}(z)) ?
The first d/(dz)(2^zlog_{2}(z)) is 2^zln(z)+(2^z)/(ln(2)z)