# Chain rule - what am i doing wrong?

#### 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.

#### harish21

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.

• machack

#### AllanCuz

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

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

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

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

• machack

#### machack

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

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