Differentiate the function: T(z) = 2z (log2z)

[Strat]

Differentiate the function: T(z) = 2^z (log₂z)

Here's what I got so far:

Applying the product rule:
(2^z)(1/zln2) + (log₂z)(2^zln2)

(2^z / zln2) + (log₂z)(2^zln2)

The answer is supposed to be 2^z[(1/zln2) + lnz] so I'm not sure where the log went or how they got to this answer. I've tried solving this different ways but this is the closest to the answer I've gotten. What am I doing wrong?

 

[ksdhart2]

Everything you've done so far is correct, barring one simple correction. You've done the calculations correctly, but what you wrote down in your first step wrong. What you wrote "(2z)(1/zln2)" is equivalent to:

$\displaystyle 2^z \cdot \dfrac{1}{z} \cdot ln(2)$

However, that's not the correct derivative. To be correct you need to use grouping symbols, like so: 2z * 1/(zln(2)). Be sure you understand why those grouping symbols make the expression what you want:

$\displaystyle 2^z \cdot \dfrac{1}{z \cdot ln(2)}$

That said, a good next step would be to apply some known rules of logarithms and simplify things. Here's what you've gotten so far:

$\displaystyle \dfrac{2^z}{z \cdot ln(2)} + log_2(z) \cdot 2^z \cdot ln(2)$

Now, let's use a rule of logarithms, $\displaystyle log_n(z)=\dfrac{ln(z)}{ln(n)}$. So, making the appropriate substitution leaves:

$\displaystyle \dfrac{2^z}{z \cdot ln(2)} + \dfrac{ln(z)}{ln(2)} \cdot 2^z \cdot ln(2)$

Now try continuing from here, by cancelling any terms as appropriate, and factoring out other common terms. You should end up with the given answer.

 

[Strat]

Thanks, that helped.