# Thread: Chain rule - what am i doing wrong?

1. ## Chain rule - what am i doing wrong?

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.

2. Originally Posted by machack
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.

3. Originally Posted by machack
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

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

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

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

4. Ah I just realized what i did wrong (thought a^x could be differentiated into x*a^(x-1))