Qualification > Math
urgent:trigonometric equation help and a few other questions
lhl_anecar:
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??
ashish:
--- Quote from: lhl_anecar 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??
--- End quote ---
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
Deadly_king:
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
ashish:
--- Quote from: Deadly_king 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
--- End quote ---
ok ok then the whole question changes :D
and thanks ;D
Deadly_king:
--- Quote from: ashish on October 21, 2010, 05:50:30 am ---ok ok then the whole question changes :D
and thanks ;D
--- End quote ---
Yupz :D
Anytime ;)
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