Chain rule - what am i doing wrong?

Mar 2010
22
1
I want to take the derivative of \(\displaystyle \left(\frac{1976}{x}\right)^x \)

Using the chain rule, i got \(\displaystyle \left(\frac{-1976}{x}\right)\left(\frac{1976}{x}\right)^{x-1} \)
But I know this is an increasing function (so the derivative should not be negative) and so I am kind of confused. Please help.
 
Feb 2010
1,036
386
Dirty South
I want to take the derivative of \(\displaystyle \left(\frac{1976}{x}\right)^x \)

Using the chain rule, i got \(\displaystyle \left(\frac{-1976}{x}\right)\left(\frac{1976}{x}\right)^{x-1} \)
But I know this is an increasing function (so the derivative should not be negative) and so I am kind of confused. Please help.
Not as easy as it seems. Check here. Click on Show Steps.
 
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Apr 2010
384
153
Canada
I want to take the derivative of \(\displaystyle \left(\frac{1976}{x}\right)^x \)

Using the chain rule, i got \(\displaystyle \left(\frac{-1976}{x}\right)\left(\frac{1976}{x}\right)^{x-1} \)
But I know this is an increasing function (so the derivative should not be negative) and so I am kind of confused. Please help.
hmmm..

let \(\displaystyle y = (\frac{1976}{x})^x \)

\(\displaystyle lny = xln( \frac{1976}{x} ) = xln(1976) -xlnx \)

Implicitly differentiate

\(\displaystyle \frac{1}{y} y^{ \prime} = ln(1976) - (lnx + 1) \)

\(\displaystyle y^{ \prime } = y[ln(1976) - (lnx + 1) ] \)

\(\displaystyle y^{ \prime } = (\frac{1976}{x})^x [ln(1976) - (lnx + 1) ] \)
 
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Mar 2010
22
1
Ah I just realized what i did wrong (thought a^x could be differentiated into x*a^(x-1))
 
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