1. ## Differentiation

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

2. Originally Posted by wahhdoe
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 $\displaystyle a^x$

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

$\displaystyle \frac{d}{dx}3^x=3^xln3$

3. Originally Posted by wahhdoe
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 $\displaystyle {dy\over dx}=y\ln 3=3^x\ln 3$.

4. Originally Posted by wahhdoe
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

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$

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

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