IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: lhl_anecar on October 16, 2010, 06:26:18 pm
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Can someone please kindly help me solve these questions
1. state the standard identity relating cos and sin arising from Pythagoras Theorem
sec^2(x) + cosec^2(x) = sec^2(x) x cosec^2(x)
I have no idea what the question wants.
and this question
2. ln x + ln(1-x)-2ln(1+x)
It looks so simple but i just cant seem to get it right.. help :(
3. 5 x (1.5)^a = 100
how do i get a?
4. find the deriavative of the following
a) y=x^5e^x + 3xlnx
b. R=(e^t+1)(t-ln t)
y=(x^5)(e^x) + 3xlnx how do i solve this??
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what do you have to do
simplify ?
or equate ?
if we simplify
its coming
ln[ (x-x*) / (1+x)* ]
x* means x squared
and if we equate
then you remove the ln
by making the other side the power of e
and equation it to the thing iside the ln bracket
that will make a quadratic equation in x
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just remember
if lnx - lnx
then its equal to ln [ x/x ]
and if lnx + lnx then it equals ln [x multiplied by x ]
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Can someone please kindly help me solve these questions
1. state the standard identity relating cos and sin arising from Pythagoras Theorem
sec^2(x) + cosec^2(x) = sec^2(x) x cosec^2(x)
I have no idea what the question wants.
the equation is
cosx2 + sinx2 = 1
for the second one i have to prove ?
secx is 1/cosx and cosecx is 1/sinx
from left hand side
sec2x + cosec2x = 1/cos2x + 1/sin2x
expressing as a single fraction
=(sin2x + cos2x)/ cos2x sin2x
= 1/sin2x * 1/cos2x
= cosec2x * sec2x
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Can someone please kindly help me solve these questions
It looks so simple but i just cant seem to get it right.. help :(
3. 5 x (1.5)^a = 100
how do i get a?
3.5 * (1.5)a =100
1.5a= 100/3.5
taking ln on both side
ln1.5a= ln(100/3.5)
power becomes multiplier (one property of log)
a=(ln(100/3.5))/ln1.5
a =8.268
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Can someone please kindly help me solve these questions
4. find the deriavative of the following
a) y=x^5e^x + 3xlnx
b. R(e^t+1)(t-ln t)
4.a
let do it part by part
y=X5ex
dy/dx = X5ex-1 *5ex
second part using quotient rule
Udv/dx + Vdu/dx
= 3x*(1/x) + 3lnx
=3+3lnx
add these two
5exX5ex-1 + 3 + 3lnx
i just hope i didn't do any mistake with superscript
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for 4b i think you missed something ???
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thanks for the help!! appreciate it loads.. it's really kind of you guys to help me out.
i just have one last question that i am still unsure about and hope u guys can help me out
which is
using the product rule to find the derivatives of the following..
a) x^2[?x + (1/?x)]
thanks again!!
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for 4b i think you missed something ???
R=(e^t+1)(t-lnt)
i missed out the = sign, sorry about that,i just don't seem to get how to differentiate ln. please help. thanks!!! :)
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oops sorry hey i have to go to school for exams ,, :-\ when i will do it when i will be back or maybe someone here will do it in the mean time ::)
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thanks for the help!! appreciate it loads.. it's really kind of you guys to help me out.
i just have one last question that i am still unsure about and hope u guys can help me out
which is
using the product rule to find the derivatives of the following..
a) x^2[?x + (1/?x)]
thanks again!!
First you need to expand the brackets and replace square roots by power 1/2.
Let y = x2(x1/2 + 1/x1/2)
Since they all have the same base(x), multiplication implies addition of powers while division implies subtraction of powers ;)
y = x(2+1/2) + x(2-1/2) ----> y = x5/2 + x3/2
From here it's easy to differentiate : dy/dx = 5/2 x3/2 + 3/2 x1/2
The required solution can be simplifies as (5x3/2 + 3x1/2)/2
Hope it helps :)
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R=(e^t+1)(t-lnt)
i missed out the = sign, sorry about that,i just don't seem to get how to differentiate ln. please help. thanks!!! :)
Is it R = (e(t+1))(t - lnt) or R = (et + 1)(t - lnt) ???
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Is it R = (e(t+1))(t - lnt) or R = (et + 1)(t - lnt) ???
R = (et + 1)(t - lnt) it is this one.. ;D sorry about the typing error. thanks!!!
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well we can use the product rule
u = e^t + 1
u' = e^t
v = t - ln t
v' = 1 - 1/t
dR/dt = u'v + uv'
= e^t (t - ln t) + (e^t +1).(1-1/t)
Tell me if this is wrong.
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y=(x^5)(e^x) + 3xlnx
how do i solve this?? i made a mistake typing the previous question... my bad, made the whole question more difficult..
someone..please help me out, again.. thanks!! thank you very very much......
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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??
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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
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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
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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
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ok ok then the whole question changes :D
and thanks ;D
Yupz :D
Anytime ;)