Author Topic: urgent:trigonometric equation help and a few other questions  (Read 2334 times)

Offline lhl_anecar

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Re: urgent:trigonometric equation help and a few other questions
« Reply #15 on: October 21, 2010, 01:18:32 am »
once again, thanks for all u guys who had helped me out, couldnt have done it without ur help..
i just need to clarify this..

y=x^5e^x + 3xlnx

how do i differentiate lnx? and how do i do the whole thing with product rule??

Offline ashish

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Re: urgent:trigonometric equation help and a few other questions
« Reply #16 on: October 21, 2010, 04:52:52 am »
once again, thanks for all u guys who had helped me out, couldnt have done it without ur help..
i just need to clarify this..

y=x^5e^x + 3xlnx

how do i differentiate lnx? and how do i do the whole thing with product rule??

d(lnx)/dx is 1/x *1
                                             =1/x

the one is the differential of x

let take another example
d(ln5x)/dx]  = (1/5x)*5
                                       = 1/x

in general d(lnf(x)) /dx  is (1/f(x)) * d(fx)/dx


PRODUCT rule is use when you have two products ..

for example

Xlnx is a product

take one term as u and the other as v
u=X
du/dx = 1

v= lnx
dv/dx= 1/x


then use this formula
dy/dx = Udv/dx + Vdu/dx
                                       = x(1/x) + (lnx)*1
                                       =1+lnx

y=x5ex + 3xlnx


y=x5ex
note: there is no product here so no product rule

dy/dx= 5exX5ex-1--------------------1

3xlnx is a product
take u as 3x
du/dx = 3

take v as lnx
dv/dx=(1/x)*1
                                      = 1/x

now us the formula
dy/dx = Udv/dx + Vdu/dx
                                        = (3x)*(1/x) +(lnx)*3
                                        = 3+3lnx---------------2
now as these two


5exX5ex-1+ 3+3lnx

Offline Deadly_king

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Re: urgent:trigonometric equation help and a few other questions
« Reply #17 on: October 21, 2010, 05:34:38 am »
Hey buddy........I think the question is as follows:

y = x5ex + 3xln x

Differentiation of ln x is always equal to 1/x.

You need to do this question step by step. Firstly, using the product rule differentiate x5ex.

Then, using the product rule again, differentiate 3xln x.

Add the two results you obtain ;)

By the way nice try ashish :D
+rep

Offline ashish

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Re: urgent:trigonometric equation help and a few other questions
« Reply #18 on: October 21, 2010, 05:50:30 am »
Hey buddy........I think the question is as follows:

y = x5ex + 3xln x

Differentiation of ln x is always equal to 1/x.

You need to do this question step by step. Firstly, using the product rule differentiate x5ex.

Then, using the product rule again, differentiate 3xln x.

Add the two results you obtain ;)

By the way nice try ashish :D
+rep

ok ok then the whole question changes :D
and thanks   ;D

Offline Deadly_king

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Re: urgent:trigonometric equation help and a few other questions
« Reply #19 on: October 21, 2010, 06:04:58 am »
ok ok then the whole question changes :D
and thanks   ;D

Yupz :D

Anytime ;)