Differentiation

Nov 2009
18
0
Hi I dont have an actual question but if I wanted to how can i differentiate a function like y=3^(x) with respect to x?

thanks
 
Dec 2009
3,120
1,342
Hi I dont have an actual question but if I wanted to how can i differentiate a function like y=3^(x) with respect to x?

thanks
This is of the form \(\displaystyle a^x\)

\(\displaystyle \frac{d}{dx}a^x=a^xlog_ea\)

\(\displaystyle \frac{d}{dx}3^x=3^xln3\)
 
Mar 2010
41
18
unfortunalty, im stuck in hicks state. IN.
Hi I dont have an actual question but if I wanted to how can i differentiate a function like y=3^(x) with respect to x?

thanks
If you don't want to memorize the formulas just take the log's.

\(\displaystyle y=3^x\)

becomes \(\displaystyle \ln y=x\ln 3\).

The ln 3 is a constant, so when you differentiate with respect to x you get, using the chain rule

\(\displaystyle {1\over y}{dy\over dx}=\ln 3\)

Popping the y on the other sides gives you \(\displaystyle {dy\over dx}=y\ln 3=3^x\ln 3\).
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Hi I dont have an actual question but if I wanted to how can i differentiate a function like y=3^(x) with respect to x?

thanks
Alternatively, transform it into a function of \(\displaystyle e\).


\(\displaystyle y = 3^x\)

\(\displaystyle = e^{\ln{3^x}}\)

\(\displaystyle = e^{x\ln{3}}\)

\(\displaystyle = (e^x)^{\ln{3}}\).


Let \(\displaystyle u = e^x\) so that \(\displaystyle y = u^{\ln{3}}\).


\(\displaystyle \frac{du}{dx} = e^x\)


\(\displaystyle \frac{dy}{du} = (\ln{3})u^{\ln{3} - 1}\)

\(\displaystyle = (\ln{3})(e^x)^{\ln{3} - 1}\)

\(\displaystyle = (\ln{3})e^{x\ln{3} - x}\)

\(\displaystyle = (\ln{3})(e^{x\ln{3}})(e^{-x})\)

\(\displaystyle = (\ln{3})(e^{\ln{3^x}})(e^{-x})\)

\(\displaystyle = (\ln{3})(3^x)(e^{-x})\).


Therefore

\(\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}\)

\(\displaystyle = e^x(\ln{3})(3^x)(e^{-x})\)

\(\displaystyle = (\ln{3})3^x\).
 
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