# Finding the derivatives of functions

1. Sep 29, 2008

### lamerali

Hi, i'm working with finding the derivatives of functions, which i'm not very comfortable with; if someone could please check my answers to the following questions i would be VERY grateful! Thank you! :)

find the derivative of the following function:

Question 1:

y = $$\frac{ 2^{x} }{ e^{x} }$$

y1 = $$\frac{ e^{x} . ln2 . 2^{x} + 2^{x} . e^{x} }{ e^{x}^{2} }$$
= $$\frac{ 2^{x} (ln2 + 1) }{ e^{x} }$$

Question 2:

f(x) = 2x ln(x$$^{2}$$ + 5)

f $$^{1}$$ (x) = 2ln(x$$^{2}$$ + 5) + (2x) . $$\frac{1}{x^{2} + 5}$$ . (2x)

= 2 ln(x$$^{2}$$ + 5) + $$\frac{4x^{2}}{x^{2} + 5}$$

Question 3:

g(x) = $$\frac{ln x}{e^{x}^{2} + 2}$$

g$$^{1}$$(x) = $$\frac{(e^{x}^{2} + 2) . (1/x) - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}$$

= $$\frac{\frac{e^{x}^{2} + 2}{x} - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}$$

for the last two questions i'm not sure if i simplified enough...if anyone could guide me in the right direction where needed i'd really appreciate it! thanks in advance!

2. Sep 29, 2008

### HallsofIvy

Staff Emeritus
Almost right you need "-" , not "+" in the numerator- quotient rule: (u/v)'= (u'v- uv')/v^2.

Yes, that looks good.

The third problem also looks good to me.

3. Sep 29, 2008

### lamerali

Great! Thank you HallsofIvy! :D