What am I doing wrong in this derivative

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my derivative of log4(r)*log2(r)= log2(r)/rln(4)+log4(r)/rln(2)

my book says the derivative should be 2(ln(r))/(r(ln(2))(ln(4))
what am I doing wrong?
 
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adelin said:
my derivative of log4(r)*log2(r)= log2(r)/rln(4)+log4(r)/rln(2)

my book says the derivative should be 2(ln(r))/(r(ln(2))(ln(4))
what am I doing wrong?

Do you know how to convert logs to different bases?
 
adelin said:
my derivative of log4(r)*log2(r)= log2(r)/rln(4)+log4(r)/rln(2)

my book says the derivative should be 2(ln(r))/(r(ln(2))(ln(4))
what am I doing wrong?

What does log4(r) mean? Is it ##\log(4r)##, log to base 4 of r, or is it ##\log(r)^4##?
 
adelin said:
my derivative of log4(r)*log2(r)= log2(r)/rln(4)+log4(r)/rln(2)

my book says the derivative should be 2(ln(r))/(r(ln(2))(ln(4))
what am I doing wrong?

You can simplify your expression further:
<br /> \frac{\log_2(r)}{r \ln 4} + \frac{\log_4(r)}{r \ln 2} = <br /> \frac{\ln r}{(r \ln 4) \ln 2} + \frac{\ln r}{(r \ln 2) \ln 4}<br /> = \frac{2 \ln r}{r \ln 2 \ln 4}<br />
 
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1/ln(b)*1/r* dy/dx(r)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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