Find the rate of change of the frequency when D, L, σ and T are varied singly. I'm reading Calculus made easy to learn the notation (I know derivatives with the limit/prime style) and also some integral calculus which I haven't seen at school yet. You can check it here: https://www.gutenberg.org/files/33283/33283-pdf.pdf

$\color{red}{\texttt{page 32}}$


  (9) The frequency n of vibration of a string of diameter D, length L
  and specific gravity σ, stretched with a force T, is given by 
  
  
    $$n = \frac{1}{DL} * \left( \frac{gT }{ πσ} \right)^{1/2}$$
  
  
  Find the rate of change of the frequency when D, L, σ and T are varied
  singly


The answers are on $\color{red}{\texttt{page 252}}$. I know you have to treat everything as a constant except for what you are differentiating at the moment. For dn/dD you only work with D and every other letter is constant, for example. 

My answers for dn/dD and dn/dL match the ones in the book... however, when I have to 'differentiate singly' stuff inside the square root I just can't get the algebra right (dn/dT and dn/dσ).
I post the derivative w.r.t T. You can first seperate the variable T from the others:

$n=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \sqrt{T}$

Then you write $\sqrt T$ as $T^{1/2}$

$n=\underbrace{\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}}_{\large{\begin{matrix}   \texttt{constant in case of} \\  \texttt{ differentiation w.r.t. T}   \end{matrix}}}\cdot T^{1/2}$

Therefore you need the derivative of $T^{1/2}$

$\frac{dn}{dT}=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \frac{1}{2}\cdot T^{1/2-1}=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \frac{1}{2}\cdot T^{-1/2}$

And $T^{-1/2}=\frac{1}{\sqrt{T}}=\sqrt{\frac{1}{T}}$

Now you can put all the terms under the radical together.

$\frac{dn}{dT}=\frac{1}{DL}\cdot \sqrt{\frac{g}{T\pi \sigma}}\cdot \frac{1}{2}=\frac{dn}{dT}=\frac{1}{2DL}\cdot \sqrt{\frac{g}{T\pi \sigma}}$

You can try the other derivative for yourself.
Recall $\sqrt{gT} = \sqrt{g} * \sqrt{T}$. 
As the book indicates, when calculating partial derivatives, we let only one variable change while the others are constant. 

Say you're looking for $\frac{dn}{dT}$, then think of $n$ as 

$n = constant * \sqrt{T}$. 

Now calculating $\frac{dn}{dT}$ should be more clear. $\frac{dn}{d\sigma}$ follows the same method.