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# Differential of 6^x? Watch

1. Is it just 6^x?
2. (Original post by mil88)
Is it just 6^x?
Think of it in exponential form.
3. (Original post by zetamcfc)
Think of it in exponential form.
Do you mean e^x?
4. (Original post by mil88)
Do you mean e^x?
Yes so how would you write 6^x in the form e^ax ?
5. (Original post by zetamcfc)
Yes so how would you write 6^x in the form e^ax ?
I'm slightly confused.

Both 6 and e are constants and therefore if it's 6 or e, it shouldn't really matter should it? When differentiating e^x, you differentiate the power and then bring the power down, and that's what I was doing for 6^x.
6. (Original post by mil88)
I'm slightly confused.

Both 6 and e are constants and therefore if it's 6 or e, it shouldn't really matter should it? When differentiating e^x, you differentiate the power and then bring the power down, and that's what I was doing for 6^x.
Ok, differentiating 6^x =/= 6^x, only e has this property. Which is why you need to get it into that form, e^ax.
7. (Original post by zetamcfc)
Ok, differentiating 6^x =/= 6^x, only e has this property. Which is why you need to get it into that form, e^ax.
Only e has that property? Oh, I never knew that!

Ok so could we do: 6^x = e^ax ------> xln6 = ax (and therefore a= ln6)

So would it be e^(ln6)x?
8. (Original post by mil88)
I'm slightly confused.

Both 6 and e are constants and therefore if it's 6 or e, it shouldn't really matter should it? When differentiating e^x, you differentiate the power and then bring the power down, and that's what I was doing for 6^x.
No. and are different functions, so they will have different slopes. They are closely related functions though. Note that:

1) since logs and exponentials are inverse functions, we have

2) if we put we have:

with
9. (Original post by mil88)
Only e has that property? Oh, I never knew that!

Ok so could we do: 6^x = e^ax ------> xln6 = ax (and therefore a= ln6)

So would it be e^(ln6)x?

10. Alternatively you could let y=a^x and take natural logs of both sides and differentiate impicitly to show that if y=a^x, dy/dx=(a^x)lna.
So d/dx(6^x) =(6^x)ln6.
11. (Original post by mil88)
Only e has that property? Oh, I never knew that!

Ok so could we do: 6^x = e^ax ------> xln6 = ax (and therefore a= ln6)

So would it be e^(ln6)x?
If what you mean is then yes! Now just simplify using the chain rule.
12. (Original post by Louisb19)
If what you mean is then yes! Now just simplify using the chain rule.
Yeah thanks I know the differentiating part, I just wasn't sure about the conversion to e form but the other users helped me!
13. (Original post by B_9710)
Alternatively you could let y=a^x and take natural logs of both sides and differentiate impicitly to show that if y=a^x, dy/dx=(a^x)lna.
So d/dx(6^x) =(6^x)ln6.
Thanks
14. (Original post by mil88)
Yeah thanks I know the differentiating part, I just wasn't sure about the conversion to e form but the other users helped me!
The idea just incase you were wondering was that

so for f(x) = 6^x gives
15. Isn't the answer just 6^xln(6)?

6^x
d/dx = 6^xln(6)
16. (Original post by atsruser)
No. and are different functions, so they will have different slopes. They are closely related functions though. Note that:

1) since logs and exponentials are inverse functions, we have

2) if we put we have:

with
Thanks but still dont understand this. What can I do with the info that a = log_e 6? where do I go from there in terms of differentiating
17. (Original post by sunburner)
Thanks but still dont understand this. What can I do with the info that a = log_e 6? where do I go from there in terms of differentiating
From there, it's just the chain rule. If and then

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