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|dw:1337454011771:dw|

It is not easy to find out.!

a = http://www.wolframalpha.com/input/?i=d%2Fdx+Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29+%2C+x%3D1

experimentX can you give me a way to your answer?

the left hand derivative must be equal to right hand derivative at x=1

for b, the function must be continuous ... so equate left hand limit to right hand limit at x=1

the right hand derivative is just "a" and left hand derivative can be found using quotient rule

I know but how can you reach to -1/6 ?

wolfram is pretty handy ... :D

yeap,but I didn't find any ways in there.

|dw:1337455871084:dw|

Ok find it out?it gets so complexity.Doesn't it?

\[ \lim_{x->1}ax+b = \lim_{x->1}\frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}}\]

i prefer to find out by using wolfram first .. then use hand to verity it.

http://www.wolframalpha.com/input/?i=lim+x-%3E1+Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29

in right hand side the limit is going to be 1 .then a+b=1
but give me a proof for a=-1/6

please show me -1/6 .
I didn't find it in wolfram...!Did you see any?

sorry - I meant take the limit as x tends to 1

Ok copy past the answer and find it out for x=1.

That was wrong site you posted.

I have never found it yet.so could you please copy past the way (lim x->1)

only i looked the answer -1/6
and derivation of function.

there is no calculation i chose step but there is only answer.

I accepted the conclusion but show me how?

\[ \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \]
I mean find this first.

ok find out this limit|dw:1337458049743:dw|

yeah, that would be -1/6

using L'hospital's rule

ok show me.i really appreciate

are you having trouble with limits?

which part .. first or last??

no but this is not easy to find out.check it out.

Ah sorry ... we need only right hand limit

both of them to each other you should take limit,because each goes to infinity !

ah .. yes ... looks like i'm nuts

im waiting yet.

yes it is going to 0/0 so what is the rest?

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Ahh ... looks like you got it!!
good night!!

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