Differential help please (1)

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nil101

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Oct 16, 2005
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Hi

Can you help me to understand how to differentiate a function with respect to x?

I have 3 problems that I am stuck on and would be thankful for any light that can be shed.

I will list them separately to avoid confusion.

Thanks

Differentiate wrt(x)

\(\displaystyle y = \frac{{\sqrt {x + 1} }}{{\ln x}}\)


using the quotient rule:
\(\displaystyle u = (x + 1)^{{\textstyle{1 \over 2}}}\)

\(\displaystyle u' = {\textstyle{1 \over 2}}(x + 1)^{^{{\textstyle{{ - 1} \over 2}}} }\)

\(\displaystyle v = \ln x\)

\(\displaystyle v' = \frac{1}{x}\)


\(\displaystyle \frac{{dy}}{{dx}} = \frac{{{\textstyle{1 \over 2}}(x + 1)^{^{{\textstyle{{ - 1} \over 2}}} } .\ln x - (x + 1)^{{\textstyle{1 \over 2}}} .\frac{1}{x}}}{{(\ln x)^2 }}\)

\(\displaystyle = \frac{{x\ln x - (2)(x + 1)^{{\textstyle{1 \over 2}}} }}{{2x(x + 1)^{{\textstyle{1 \over 2}}} (\ln x)^2 }}\)

\(\displaystyle = \frac{{x\ln x - 2\sqrt {x + 1} }}{{2x\sqrt {x + 1} .(\ln x)^2 }}\)... this was my answer but ...


I know the answer is... \(\displaystyle \frac{{x\ln x - 2(x + 1)}}{{2x\sqrt {x + 1} .(\ln x)^2 }}\)

I dont know how to get rid of the square root in the numerator.

Please can you help?

Thanks
 
Upon using the quotient rule, you end up with:

\(\displaystyle \L\\\frac{\frac{ln(x)}{2\sqrt{x+1}}-\frac{\sqrt{x+1}}{x}}{(ln(x))^{2}}\)

You can just cross multiply the numerator to eliminate the radicals.

\(\displaystyle \L\\\frac{xln(x)-2(x+1)}{2x\sqrt{x+1}(ln(x))^{2}}\)
 
nil101 said:
Hi

Can you help me to understand how to differentiate a function with respect to x?

I have 3 problems that I am stuck on and would be thankful for any light that can be shed.

I will list them separately to avoid confusion.

Thanks

Differentiate wrt(x)

\(\displaystyle y = \frac{{\sqrt {x + 1} }}{{\ln x}}\)


using the quotient rule:
\(\displaystyle u = (x + 1)^{{\textstyle{1 \over 2}}}\)

\(\displaystyle u' = {\textstyle{1 \over 2}}(x + 1)^{^{{\textstyle{{ - 1} \over 2}}} }\)

\(\displaystyle v = \ln x\)

\(\displaystyle v' = \frac{1}{x}\)


\(\displaystyle \frac{{dy}}{{dx}} = \frac{{{\textstyle{1 \over 2}}(x + 1)^{^{{\textstyle{{ - 1} \over 2}}} } .\ln x - (x + 1)^{{\textstyle{1 \over 2}}} .\frac{1}{x}}}{{(\ln x)^2 }}\)

\(\displaystyle = \frac{{x\ln x - (2)(x + 1)^{{\textstyle{1 \over 2}}} }}{{2x(x + 1)^{{\textstyle{1 \over 2}}} (\ln x)^2 }}\)

\(\displaystyle = \frac{{x\ln x - 2\sqrt {x + 1} }}{{2x\sqrt {x + 1} .(\ln x)^2 }}\)... this was my answer but ...


I know the answer is... \(\displaystyle \frac{{x\ln x - 2(x + 1)}}{{2x\sqrt {x + 1} .(\ln x)^2 }}\)

I dont know how to get rid of the square root in the numerator.

Please can you help?

Thanks
Click to expand...

Before you get too far into the algebra, factor as much as possible.

Note:
\(\displaystyle \frac{{dy}}{{dx}} = \frac{{{\textstyle{1 \over 2}}(x + 1)^{^{{\textstyle{{ - 1} \over 2}}} } .\ln x - (x + 1)^{{\textstyle{1 \over 2}}} .\frac{1}{x}}}{{(\ln x)^2 }} = \frac{(x+1)^{\frac{-1}{2}}(\frac{1}{2}lnx - (x+1)\frac{1}{x}}{(lnx)^2})\)
 
Hello, nil101!

Differentiate: \(\displaystyle \L\,y\;=\;\frac{\sqrt{x + 1}}{\ln x}\)

\(\displaystyle \L\;\;\frac{dy}{dx}\;=\;\frac{\frac{1}{2}(x\,+\,1)^{-\frac{1}{2}}\cdot\ln x\,-\,(x\,+\,)^{\frac{1}{2}}\cdot\frac{1}{x}}{(\ln x)^2}\;\;\) . . . Correct!
Click to expand...
This is a complex fraction . . . It has more than two "levels".

Look at what we have: \(\displaystyle \L\;\frac{\frac{1}{2}\cdot\frac{1}{\sqrt{x\,+\,1}}\cdot\ln x\;-\;\frac{1}{x}\cdot\sqrt{x\,+\,1}}{(\ln x)^2}\)

It's "little" denominators that are annoying. \(\displaystyle \;\)Find their LCD.


Multiply top and bottom by \(\displaystyle 2x\sqrt{x\,+\,1}:\)

\(\displaystyle \L\;\;\frac{2x\sqrt{x\,+\,1}}{2x\sqrt{x\,+\,1}}\,\cdot\, \frac{\frac{1}{2}\cdot\frac{1}{\sqrt{x\.+\,1}}\cdot\ln x\:-\:\frac{1}{x}\cdot\sqrt{x\,+\,1}}{(\ln x)^2}\)

\(\displaystyle \L\;\;= \;\frac{\not{2}x\sout{\sqrt{x\,+\,1}}\,\cdot\,\frac{1}{\not{2}}\cdot\frac{1}{\sout{\sqrt{x\,+\,1}}}\cdot\ln x \:- \:2\not{x}\sqrt{x\,+\,1}\,\cdot\,\frac{1}{\not{x}}\cdot\sqrt{x\,+\,1}}{2x\sqrt{x\,+\,1}\cdot(\ln x)^2}\)

And we have: \(\displaystyle \L\:\frac{x\cdot\ln x \,- \,2(x\,+\,1)}{2x\cdot\sqrt{x\,+\,1}\cdot(\ln x)^2}\)
 
Thank you to all of you guys for helping on this one, particularly Soroban, as I'm having difficulty with complex fractions.

:)
 
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