Logarithmic differentiation Question

[Jto_012]

Hello

I am not clear how do i get the value of in 2 from (x in 2)'. Where did the X goes? What is the derivative of In 2?

Hope you can help on this.

 

[Harry_the_cat]

(ln 2) is a constant. The derivative of (x times a constant) is just the constant.

Another (easier to understand) way to differentiate \(\displaystyle f(x)=2^x\) is to observe that \(\displaystyle 2= e^{ln 2}\).

So, \(\displaystyle f(x) = 2^x = e^{x (ln 2)}\)

which leads to

\(\displaystyle f '(x) = (ln 2) e^{x (ln 2)} = (ln 2) 2^x\).

 

[Steven G]

Where does the x go in taking the derivative of 3x? Note that the derivative of 3x is 3.

ln2 is as much of a number as 17. x*17 = 17*x, x*ln2 = ln2*x.

 

[pka]

Hello
I am not clear how do i get the value of in 2 from (x in 2)'. Where did the X goes? What is the derivative of In 2?
Hope you can help on this.

We want the derivative of [imath]2^x[/imath] using logarithmic differentiation.
Start with [imath]y=2^x[/imath]. The use logarithms [imath]\log(y)=\log\left(2^x\right)=x\log\left(2\right)[/imath].
Using implicit derivatives we get, [imath]\dfrac{y^{\prime}}{y}=\log(2)[/imath].
Thus [imath]y^{\prime}=y\log(2)=2^x\log(2)[/imath].
That is [imath]y^{\prime}=2^x\log(2)[/imath]


[imath][/imath][imath][/imath][imath][/imath]

 

[Dr.Peterson]

Hello

I am not clear how do i get the value of in 2 from (x in 2)'. Where did the X goes? What is the derivative of In 2?

You typed the function name as "in" and "In", suggesting that you are one of the many who misread "ln" ("ell en", for "Logarithm Natural") as "In" ("eye en"):

I am not clear how do i get the value of in 2 from (x in 2)'. Where did the X goes? What is the derivative of In 2?​

I want to make sure you see it correctly!

It's hard to see in the sans serif font used here, though it should be easy enough to see the difference in your book, if you look carefully.

Others have, of course, answered the actual question.