Logarithmic differentiation

[sigma]

Can somebody help me with these 2 logarithmic diffenentiation problems? I'm sorry I don't have any work because I don't even know where to begin and I have to know these ASAP.

Find y'

\(\displaystyle \
\L\
y = \frac{{\ln \sqrt {x + 3} }}{{\sqrt {\log (x + 3)} }}
\\)

and

\(\displaystyle \
\L\
y = \ln (\cos x) \cdot \log _2 (x^2 + x)
\\)

Much appreciated.

 

[sigma]

I think I may have got the second one. Is this how you do it? This answer isn't simplified all the way but I want to make sure I'm on the right track. Still can't figure out the first one.

\(\displaystyle \
\L\
y' = \frac{1}{{\cos x}}( - \sin x) \cdot \log _2 (x^2 + x) + \ln (\cos x)\frac{1}{{(\ln 2)}} \cdot \frac{1}{{(x^2 + x)}}(2x)
\\)


\(\displaystyle \
\L\
=y' = \frac{{ - \sin x}}{{\cos x}} \cdot \log _2 (x^2 + x) + \frac{{\ln (\cos x)(2x)}}{{(\ln 2)(x^2 + x)}}
\\)

 

[galactus]

Perhaps you could start by simplifying.

Assuming base 10, Use the identity \(\displaystyle \L\\log(x+3)=\frac{ln(x+3)}{ln(10)}\)

\(\displaystyle \L\\\frac{ln(\sqrt{x+3})}{sqrt{\frac{ln(x+3)}{ln(10)}}}\)

=\(\displaystyle \L\\ln(sqrt{x+3})\frac{sqrt{ln(10)}}{sqrt{ln(x+3)}}\)

But,\(\displaystyle ln(sqrt{x+3})=\frac{1}{2}ln(x+3)\)

\(\displaystyle \L\\\frac{1}{2}\sqrt{ln(10)}\frac{ln(sqrt{x+3})}{sqrt{ln(x+3)}}\)

=\(\displaystyle \L\\\frac{1}{2}\sqrt{ln(10)}\frac{d}{dx}[\sqrt{ln(x+3)}]\)

Now, just differentiate \(\displaystyle \sqrt{ln(x+3)}\). Don't forget the chain rule.

 

[sigma]

Great thanks. Did I do the second one correctly?

 

[sigma]

No answer to my last but I think I did it right anyways.

But about the second one, couldn't I just do this? Use the quotent rule?

\(\displaystyle \
\L\
y' = \frac{{\frac{1}{{\sqrt {x + 3} }}\left( {\frac{1}{2}(x + 3)^{ - \frac{1}{2}} } \right)\left[ {\sqrt {\log (x + 3)} } \right] - \left( {\ln \sqrt {x + 3} } \right)\left( {\frac{1}{2}\left( {\log (x + 3)} \right)^{ - \frac{1}{2}} } \right)\left( {\frac{1}{{\ln 10}} \cdot \frac{1}{{x + 3}}} \right)}}{{\left[ {\sqrt {\log (x + 3)} } \right]^2 }}
\\)

 

[soroban]

Hello, sigma!

about the first one, couldn't I just do this? Use the quotent rule?

\(\displaystyle \L y'\;=\;\frac{\frac{1}{\sqrt{x\,+\,3} }\left({\frac{1}{2}(x\,+\,3)^{-\frac{1}{2}}}\right)\left[{\sqrt{\log(x\,+\.3)}}\right]\,-\,\left({\ln\sqrt{x\,+\.3}}\right)\left({\frac{1}{2}\left({\log(x\,+\,3)}\right)^{-\frac{1}{2}}}\right)\left({\frac{1}{{\ln 10}}\cdot\frac{1}{x\,+\,3}}\right)}{\left[{\sqrt{\log(x\,+\,3)}} \right]^2}\)

Yes, you certainly can!

In you second answer, the last factor should be: \(\displaystyle (2x\,+\,1)\)

 

[sigma]

Great thanks!