# Simple derivatives help...

• Mar 31st 2009, 08:23 PM
yekkalu
Simple derivatives help...
Hi....I am trying to learn derivatives my self....I am able to do the simple ones........but finding difficulty in solving the below problems...Can anybody please help me.(Bow)

Derivatives of y=x\(lnx), y=squareroot(sinx+cosx), y=1\(1-x), y=(4x)power x.

I once again request.....pls help.
• Mar 31st 2009, 08:38 PM
coolguy99
for the first one..

y = x/ln(x)

alright, so, quotient rule states:

(u'v - v'u)/(v^2)

let u = x, and v = ln(x)

and keep in mind that the derivative of ln(x) = 1/x

give that a shot :)

for the second one,

square root (sin(x) + cos(x)) can be written as..

(sinx + cosx)^(1/2)

using the chain rule,

(1/2)[(sinx + cosx)^(-1/2)] (cos(x) - sin(x))

keeping in mind that the derivative of sinx = cosx, and the derivative of cosx = -sinx.

the third one I'm not really sure what you meant by it, so, try to clarify hehe
• Mar 31st 2009, 08:50 PM
yekkalu
hey thanks...its so quick and solutions are simple....

third one is y = 1/(1-x)

fourth one is y = (4x)^x .Pls help.
• Mar 31st 2009, 09:12 PM
coolguy99
alright so,

y = 1/(1-x)

you can do this two ways. Quotient rule, as described above, or, it can also be written as

y = (1-x)^(-1), just using basic algebra

so, using the chain rule,

y' = (-1)(-1)(1-x)^-2,
y' = (1-x)^(-2)

or

y' = (1)/[(1-x)^2]

noting that the extra (-1) comes from taking the derivative of what is inside the parenthesis, and multiplying it to the whole thing.

y = (4x)^x

this is a little bit harder...

take the natural log of both sides,
ln(y) = ln (4x)^x
do a little algebra magic,
ln(y) = x ln(4x)
derivative of both sides..
y'/y = ln(4x) + 1
multiply by y
y' = (y)ln(4x) + y

y was originally = (4x)^x, so, plugging back in

y' = [(4x)^x](ln(4x)) + (4x)^x
• Apr 1st 2009, 07:57 PM
yekkalu
That was really helpful....Thanks a lot.