# Finding derivatives

1. Jun 3, 2014

### TommG

find the indicated derivative
dp/dq if p = 1/(√q+1)

my work

[(1/(√(q+h+1))) - (1/(√(q+1))] $\div$h

[((√(q+1)) - (√(q+h+1)))/((√(q+h+1))(√(q+1)))] $\div$h

[(q+1-q-h-1)/(((√q+h+1)(√q+1))((√q+1)+(√q+h+1)))]$\div$ h

[-h//(((√(q+h+1))(√(q+1)))((√(q+1))+(√(q+h+1))))]$\div$ h

-1//(((√(q+h+1))(√(q+1)))((√(q+1))+(√(q+h+1))))

-1/[((√(q+1))(√(q+1)))((√(q+1))+(√(q+1)))]

the answer in the book is -1/[2(q+1)(√(q+1))]

is my answer the same as the book or is there something else I still need to do?

Last edited: Jun 3, 2014
2. Jun 3, 2014

### Ray Vickson

Do you mean
$$p = \frac{1}{\sqrt{q+1}} \text{ or } p = \frac{1}{\sqrt{q} + 1}?$$
In text you should write the first as p = 1/√(q+1) and the second as p = 1/(1+√p) or 1/((√p)+1).

3. Jun 3, 2014

### TommG

I am sorry it is the first one $$p = \frac{1}{\sqrt{q+1}}$$

4. Jun 3, 2014

### CAF123

So your answer was $$-\frac{1}{(q+1)(2\sqrt{q+1})}$$ and the book's was $$-\frac{1}{2(q+1)(\sqrt{q+1})}?$$

If so, yes they are the same. q is just a variable. It is the norm to see, for example, 2x rather than x2. And the latter becomes problematic when dealing with a product of different variables.