Differentiating $2^{n/100}$ using the chain rule

Question

I am trying to find the derivative of $2^{n/100}$ with respect to $n$. I know that I have to use the chain rule to differentiate this.

I have:

$$f(n) = 2^{n/100} = e^{\ln2 \cdot n/100}$$

For my $u$, I have set it to $\ln2 \cdot (n/100)$.

$$\begin{align} df/du &= e^{ln2 \cdot n/100}\\ du/dn &= \ln2/100\\ df/dn &= e^{\ln2 \cdot n/100} \cdot \ln2/100 \end{align}$$

However, I checked the answer and the derivative is supposed to be:

$$df/dn = 1/25 \cdot 2^{n/100 - 2} \cdot \ln2$$

I am not sure how the above answer was derived. Any insights are appreciated.

Answer 1

It's merely a simplification thing. We have, in your answer,

$$e^{\ln2\cdot n/100}\cdot\frac{\ln2}{100}$$

First, we note that $e^{\ln(2)} = 2$. Then this becomes

$$2^{n/100}\cdot\frac{\ln2}{100}$$

Next, we note that $100 = 2^2 \cdot 25$. Thus,

$$2^{n/100}\cdot\frac{\ln2}{100} = 2^{n/100}\cdot\frac{\ln2}{25} \cdot \frac{1}{2^2}$$

We can thus subtract $2$ from the exponent of the $2^{n/100}$. Thus, we obtain

$$ 2^{n/100 - 2}\cdot\frac{\ln2}{25}$$

the answer you were given.

Answer 2

$$1/25 \cdot 2^{n/100 - 2} \cdot \ln2=1/25 \cdot 2^{-2} \cdot 2^{n/100} \cdot \ln2=e^{\ln 2 \cdot n/100} \cdot \ln2/100$$

So your answer is the same as the standard answer.

Answer 3

This is a case where logarithmic differentiation makes life much easier.

Making the problem more general $$y=a ^{bn}\implies \log(y)=b \log(a)\,n$$ Diffentiate both sides $$\frac {y'}y=b \log(a)\implies y'=b \log(a)\,a^{bn}$$