Question

derivative of 5^(1/-x)

Answer 1

I looked it up and I don't understand where the ln comes from.

Answer 2

\(\large \frac{d}{dx}a^x=a^xln(a)\)

Answer 3

in this case we have \(\huge 5^{\frac{1}{-x}} => 5^{\frac{1}{-x}}*ln(5)*\frac{d}{dx}\frac{1}{-x}\)

Answer 4

the 3rd term comes in because of the chain rule

Answer 5

So if we had 2^(sinpiex) the derivative would be 2^(sinpiex) times the ln of 2?

Answer 6

I think it'd be something like \(\large 2^{sin(\pi x)\large*ln(2)*cos(\pi x)}\)

Answer 7

For \(a>0\),
\[a^x=e^{\ln a^x}=e^{x\ln a}~~\implies~~\frac{d}{dx}[e^{x\ln a}]=\ln a~e^{x\ln a}=\ln a~a^x\]

Answer 8

Or more generally,
For \(a>0\),
\[a^{f(x)}=e^{\ln a^{f(x)}}=e^{f(x)\ln a}~~\implies~~\frac{d}{dx}[e^{f(x)\ln a}]=\ln a~e^{f(x)\ln a}=\ln a~f'(x)~a^x\]

Answer 9

so when you took the derivative, why are there two ln of a ?

Answer 10

\[\frac{5^{-1/x} \log (5)}{x^2} \]