differentiating logarithms (1 Viewer)

[onebytwo]

please help to differentiate the following

(i) logx3

(ii) logx(3^x)

(iii) y = x^x

(iv) y = x^(logx)

(v) y = x^(1/x)

thanks in advance

 

[darkliight]

(i) Rewrite this one as d/dx ln(3)/ln(x). Now ln(3) is just a constant, so we have
ln(3) * d/dx 1/ln(x). Chain (or quotient) rule will finish this one up for you.

(ii) Rewrite this one too, d/dx ln(3^x)/ln(x) = ln(3) * d/dx x/ln(x). Quotient rule and you're done.

(iii) Implicit differentiation, rewrite as ln(y) = xln(x). We then have y'/y = ln(x) + 1, rearranging gives y' = y(ln(x) + 1), we know y so we can write y' = x^x(ln(x) + 1). Same deal with the next two ...

 

[onebytwo]

thanks heaps
i didnt think of using implicit diff.
i think i made progress by doing it this way:
y=x^x
lny=xlnx
so, e^(xlnx)=y
so, dy/dx=(1+x)e^(xlnx).
then its pretty much the same deal with the others
i think this method is valid, but i'll have to check to see if we arrive at the same answer.
thanks for ur precious sleep time anyway

 

[darkliight]

Your method is fine,

Two small things, just double check you differentiated e^(xln(x)) properly, and remember you know e^(xln(x)) = x^x so you can rewrite this bit if you want.

 

[Mountain.Dew]

please help to differentiate the following

(iv) y = x^(logx)

(v) y = x^(1/x)

thanks in advance

heres my 2 cents...

for (iv) y = x^(lnx) [assuming that logx = lnx]

we log both sides... lny = ln[x^(lnx)]

ln y = lnx * lnx = (lnx)^2

now, implicit diff...

(d ln y/dy) * (dy/dx) = 2(lnx)^ 1 * [1/x]

(1/y) * (dy/dx) = 2(lnx)/x

so dy/dx = 2ylnx / x = 2[lnx][x^(lnx)] / x

for (v) => y = x^(1/x)

x^(1/x) = e^([1/x][lnx])

so, we have y = e^([1/x][lnx])

therefore, dy/dx = [d e^([1/x][lnx])/d([1/x][lnx])] * d([1/x][lnx])/dx = e^([1/x][lnx]) * [-x^(-2)lnx + 1/x(1/x)]

dy/dx = x^(1/x) * [(1/x^2)(1-lnx)]

hopefully that they are right...M.D. ^^

 

[onebytwo]

theyre off the teachers work sheet, so i dont know if im doing them right.
thanks for ur time any way, guys

 

[STx]

for v) i get:

y=x^1/x
taking log of both sides:
lny=lnx^1/x=lnx/x
diff. both sides implicity:
y'/y=1/x²-lnx/x²
= 1-lnx/x²
y'=x^1/x(1-lnx)/x²

edit: got it now, made a careless when differentiating.

 

[pLuvia]

For (v) I get:
y=x^1/x
y=e^(1/x)lnx
d/dx(e^(1/x)lnx)
=(1/x*1/x+(-1/x²*lnx)[e^(1/x)lnx)]
=[1/x²-lnx/x²][e^(1/x)lnx)]
=1/x²[1-lnx][e^(1/x)lnx)]
=1/x²[1-lnx][x^1/x]

Which is the same as M.D's

 

[pLuvia]

for v) i get:

y=x^1/x
taking log of both sides:
lny=lnx^1/x=lnx/x
diff. both sides implicity:
y'/y=1/x²*lnx/x²=lnx/x⁴
Therefore, y'=ylnx/x⁴=x^1/x(lnx)/x⁴

Mines not the same as Mountain.Dew's..?

You sure you did right?

y'/y=1/x²*lnx/x²=lnx/x⁴

Shouldn't this be
y'/y=[x/x-lnx]/x²
y'=y[1-lnx]/x²
=x^1/x[1-lnx]/x²
=[1/x²][1-lnx][x^1/x]

Which is the same as M.C's

 

[conman]

Any one have any note about differentiating logarithms?

 

[Wackedupwacko]

depends on what u mean by notes. but id suggest u know ur log laws so u can easily flip ur logs around to make things easier and secondedly know the derivitive of

y= ln (f(x))

which is

y' = f'(x)/f(x)

once u know those 2 u should be able to differentiate using logs without any problem. side note log a B = ln a/ln B (which u then differentiate using quotion) change of base is the only way to differentiate a non base e log

 

[darkliight]

And by log a B = ln a/ln B you mean log a B = ln B/ln a .. right?

 

[Riviet]

Yeah, I think he meant log_ab=log_nb / log_na, where n is any base, probably base 10 or base e for easier conversion with the calculator.