1. ## differentiation

hey

i got a problem with differentiation, and im not having much luck

the question is if i have a exponential function

y=12*0.5^x

how would find y'

2. Hello,

Transform it with logarithms. You can't derivate correctly when there is x in the power.

ln(ab)=ln(a)+ln(b)

ln(a^b)=b ln(a)

;-)

3. okay thanks

so its = ln(12) +ln(0.5^x)
= ln(12) + xln(0.5)
=x ln(0.5) + ln(12)
=x ln(0.5 * 12)
=x ln(6)
= 1.79x

4. Originally Posted by Quill
okay thanks

so its = ln(12) +ln(0.5^x)
= ln(12) + xln(0.5)
=x ln(0.5) + ln(12)
=x ln(0.5 * 12)
=x ln(6)
= 1.79x
No.

$\displaystyle y = k \, (a^x) = k\, e^{\ln a^x} = k \, e^{x \ln a}$ which just has the form $\displaystyle y = k \, e^{bx}$. And you know how to differentiate that form, right?

So you end up with $\displaystyle \frac{dy}{dx} = k \, (\ln a) \, a^x$. Capisce?

A formula worth remembering.

5. um i think i got it

y=12*0.5^x
=12(0.5^x)
=12(e^ln0.5^x)
=12(e^xln0.5)
=12(ln0.5)0.5^x

thanks again

6. Originally Posted by Quill
um i think i got it

y=12*0.5^x
=12(0.5^x)
=12(e^ln0.5^x)
=12(e^xln0.5)
=12(ln0.5)0.5^x

thanks again
indeed. now that you know two ways to find the derivative of functions like this, it's time for the general rule.

$\displaystyle \frac d{dx}(a^x) = a^x \ln a$ where $\displaystyle a > 0$ is a constant

7. thanks

one last thing, when i graph it, i get an almost straight line, is that normal

8. Originally Posted by Quill
thanks

one last thing, when i graph it, i get an almost straight line, is that normal
It might look that way, depending on the scale you've used. But it most certainly is not a straight line.

9. are yer that made it work

thanks