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 $\displaystyle {dy\over dx}=y\ln 3=3^x\ln 3$.
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
$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$
$\displaystyle = e^x(\ln{3})(3^x)(e^{-x})$
$\displaystyle = (\ln{3})3^x$.