derivative of ln

[prof_fr]

can anyone help, please

what is the derivative of

ln(1/2*x)

i'm tearing my hair out

thanks

 

[pka]

can anyone help, please
what is the derivative of ln(1/2*x)

\(\displaystyle \dfrac{1}{x}\)

 

[Deleted member 4993]

ln(½ * x) = ln(½) + ln(x)

 

[srmichael]

Or use the fact that \(\displaystyle y=\ln(u) \longrightarrow y'=\frac{u'}{u}\) thus \(\displaystyle u=\frac{1}{2}x \longrightarrow u'=\frac{1}{2}\) thus \(\displaystyle y'=\frac{\frac{1}{2}}{\frac{1}{2}x}=\frac{1}{x}\)

many ways to skin that cat

 

[pka]

Or use the fact that \(\displaystyle y=\ln(u) \longrightarrow y'=\frac{u'}{u}\) thus \(\displaystyle u=\frac{1}{2}x \longrightarrow u'=\frac{1}{2}\) thus \(\displaystyle y'=\frac{\frac{1}{2}}{\frac{1}{2}x}=\frac{1}{x}\) many ways to skin that cat

I have got ask why anyone would want a student to use u-substitution to find a derivative?
I am even apposed to using u-substitution for basic integrals.

You u-substituters have got to face reality.
In less that fifteen years calculus textbooks will not have chapters on techniques of integration.
Sites like wolframalpha will make all that obsolete. That app is available on any mobile device.
If you are someone who thinks that student demands do not drive educational policy, you are delusional.
Keith Devlin wrote on this at the MAA website almost 15 years ago.

For example, if our students do not grasp at once \(\displaystyle \int {\frac{{{e^x}dx}}{{1 + {e^{2x}}}}}\) do not grasp at once that is an arctan form then you have failed to teach methods integration.
I agree whole hearty.

 

[srmichael]

I have got ask why anyone would want a student to use u-substitution to find a derivative?
I am even apposed to using u-substitution for basic integrals.

You u-substituters have got to face reality.
In less that fifteen years calculus textbooks will not have chapters on techniques of integration.
Sites like wolframalpha will make all that obsolete. That app is available on any mobile device.
If you are someone who thinks that student demands do not drive educational policy, you are delusional.
Keith Devlin wrote on this at the MAA website almost 15 years ago.

For example, if our students do not grasp at once \(\displaystyle \int {\frac{{{e^x}dx}}{{1 + {e^{2x}}}}}\) do not grasp at once that is an arctan form then you have failed to teach methods integration.
I agree whole hearty.

Unfortunately, the reality of it all is that the textbooks still explain the derivative of ln(u) as I stated above. I agree with you, though. Personally, I just tell the kids I tutor to use the log rules and make ln[1/2(x)] = ln(1/2) + ln(x) then go from there.

 

[prof_fr]

derivative of ln(1/2*x)

Thank you all so much for you replies and explanations

But could you please explain why ln(x) = 1/x AND ln(1/2*x) = 1/x

Thank you

 

[srmichael]

Thank you all so much for you replies and explanations

But could you please explain why ln(x) = 1/x AND ln(1/2*x) = 1/x

Thank you

What do you mean? ln(x) does NOT equal 1/x. The DERIVATIVE of ln(x) = 1/x. Likewise, the derviative of ln(1/2*x) also equals 1/x. I think we explained this a few times above. Where are you confused?

 

[pka]

, I just tell the kids I tutor to use the log rules and make ln[1/2(x)] = ln(1/2) + ln(x) then go from there.

That is the correct way to proceed.
If \(\displaystyle c>0\) then the derivative of \(\displaystyle \ln(cx)=\frac{1}{x}\) because \(\displaystyle \ln(cx)=\ln(c)+\ln(x)\).
\(\displaystyle \ln(c)\) is a constant. The derivative of constant is zero. That leaves only \(\displaystyle \frac{1}{x}\).

 

[prof_fr]

Thank you all so much - NOW I get it.

Sorry I did not state the derivatives of ln were what I was querying.

Perhaps between you you could write a book of definitions as this bit was nowhere to be found except with you good selves


Thanks again