- #1
kuahji
- 394
- 2
[SOLVED] derivative with logarithms
Ok, so the problem problem probably isn't as bad as I'm making it, either that or its because its getting late & my brain just isn't functioning.
Find the derivative of y with respect to r.
y=[tex]log _2 \left( r \right)[/tex] * [tex]log _4 \left( r \right)[/tex]
The first thing I thought to do was use the product rule which yielded
y'=[tex]log _2 \left( r \right)[/tex]*(1/ln2)(1/r)+[tex]log _4 \left( r \right)[/tex](1/ln4)(1/r)
I then changed the [tex]log _2 \left( r \right)[/tex] to (ln r/ln2) & did the same to the other logarithm which created a complex fraction that I condensed down to
(ln r)/(r ln 2 * ln 2) + (ln r)/(r ln 4 * ln 4)
Cross multiplying gave
[(ln r)(ln 4)^2+ (ln r)(ln 2)^2]/[r (ln 2)^2 (kn 4)^2]
Here is where I'm stuck, because the book likes (2 ln r)/(r ln 2 * ln 4).
The book also showed different steps. To begin with it shows making the logs into ln.
which gives
y=(ln r)^2/(ln 2 * ln 4)
then take the derivative, but here is where I'm a bit lost. It shows
y'=1/(ln2 *ln 4) * (2 ln r) * (1/r)
I can see where the (2 ln r) & the (1/r) come from the chain rule, but not so much the 1/(ln2 *ln 4) part.
Ok, so the problem problem probably isn't as bad as I'm making it, either that or its because its getting late & my brain just isn't functioning.
Find the derivative of y with respect to r.
y=[tex]log _2 \left( r \right)[/tex] * [tex]log _4 \left( r \right)[/tex]
The first thing I thought to do was use the product rule which yielded
y'=[tex]log _2 \left( r \right)[/tex]*(1/ln2)(1/r)+[tex]log _4 \left( r \right)[/tex](1/ln4)(1/r)
I then changed the [tex]log _2 \left( r \right)[/tex] to (ln r/ln2) & did the same to the other logarithm which created a complex fraction that I condensed down to
(ln r)/(r ln 2 * ln 2) + (ln r)/(r ln 4 * ln 4)
Cross multiplying gave
[(ln r)(ln 4)^2+ (ln r)(ln 2)^2]/[r (ln 2)^2 (kn 4)^2]
Here is where I'm stuck, because the book likes (2 ln r)/(r ln 2 * ln 4).
The book also showed different steps. To begin with it shows making the logs into ln.
which gives
y=(ln r)^2/(ln 2 * ln 4)
then take the derivative, but here is where I'm a bit lost. It shows
y'=1/(ln2 *ln 4) * (2 ln r) * (1/r)
I can see where the (2 ln r) & the (1/r) come from the chain rule, but not so much the 1/(ln2 *ln 4) part.