# Differentiation

#### 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

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 $$\displaystyle a^x$$

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

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

• HallsofIvy and wahhdoe

#### autumn

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 $$\displaystyle {dy\over dx}=y\ln 3=3^x\ln 3$$.

• HallsofIvy and matheagle

#### Prove It

MHF Helper
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

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

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

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

• HallsofIvy