Brain Stuck on differentiation for exponential function

[MrGinsu]

Hello,

I'm trying to understand what is happening with the differentiation formula for exponential functions. I know this concept is very important so I want to make sure I understand what's going on.

Find d/dx

y = 5 ^ (-1/x)

I understand how to get to 5 ^ (-1/x) ln 5 but not the rest. The book gives an answer of 5 ^ (-1/x)/x ^ 2

Can someone explain where the x^2 is coming from?

Thanks in advance,
Clayton

 

[Unco]

The difference is: \(\displaystyle \mbox{\frac{d}{dx} \left(x\right) = 1}\) whereas \(\displaystyle \mbox{\frac{d}{dx} \left(-\frac{1}{x}\right) = \frac{1}{x^2}}\).


So, if you would like an understanding, let's go the non-formula route.

\(\displaystyle \mbox{ y = 5^{\left(-\frac{1}{x}\right)}}\)

Take natural logs of both sides and simplify

\(\displaystyle \mbox{ \ln{y} = -\frac{1}{x}\ln{(5)}}\)

Differentiate both sides implicitly wrt x:

\(\displaystyle \mbox{ \frac{d}{dx} \left(\ln{y}\right) = \frac{1}{y} \frac{dy}{dx}}\)

\(\displaystyle \mbox{ \frac{d}{dx} \left(-\frac{1}{x}\ln{(5)}\right) = \frac{d}{dx} \left( -\ln{(5)}x^{-1}\right) = \ln{(5)}x^{-2} = \frac{\ln{(5)}}{x^2}}\)

So we have

\(\displaystyle \mbox{ \frac{1}{y} \frac{dy}{dx} = \frac{\ln{(5)}}{x^2}}\)

Multiplying both sides by y and substituting back \(\displaystyle \mbox{y = 5^{\left(-\frac{1}{x}\right)}\) gives

\(\displaystyle \mbox{ \frac{dy}{dx} = \frac{5^{\left(-\frac{1}{x}\right)}\ln{(5)}}{x^2}}\)

 

[MrGinsu]

The difference is: \(\displaystyle \mbox{\frac{d}{dx} \left(x\right) = 1}\) whereas \(\displaystyle \mbox{\frac{d}{dx} \left(-\frac{1}{x}\right) = \frac{1}{x^2}}\).


So, if you would like an understanding, let's go the non-formula route.

\(\displaystyle \mbox{ y = 5^{\left(-\frac{1}{x}\right)}}\)

Take natural logs of both sides and simplify

\(\displaystyle \mbox{ \ln{y} = -\frac{1}{x}\ln{(5)}}\)

Differentiate both sides implicitly wrt x:

\(\displaystyle \mbox{ \frac{d}{dx} \left(\ln{y}\right) = \frac{1}{y} \frac{dy}{dx}}\)

\(\displaystyle \mbox{ \frac{d}{dx} \left(-\frac{1}{x}\ln{(5)}\right) = \frac{d}{dx} \left( -\ln{(5)}x^{-1}\right) = \ln{(5)}x^{-2} = \frac{\ln{(5)}}{x^2}}\)

So we have

\(\displaystyle \mbox{ \frac{1}{y} \frac{dy}{dx} = \frac{\ln{(5)}}{x^2}}\)

Multiplying both sides by y and substituting back \(\displaystyle \mbox{y = 5^{\left(-\frac{1}{x}\right)}\) gives

\(\displaystyle \mbox{ \frac{dy}{dx} = \frac{5^{\left(-\frac{1}{x}\right)}\ln{(5)}}{x^2}}\)

Awesome! Your explanation helped me see what was going on.

Thanks so much, Unco.