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.

Reply 5

zetamcfc

19

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.

Reply 6

username2097061

OP

2

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?

Reply 7

atsruser

11

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.

6^x

and

e^x

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

e^{\log_e u} = u

2) if we put

u=6^x

we have:

6^x = e^{ \log_e 6^x} = e^{x \log_e 6} = e^{ax}

with

a=\log_e 6

Reply 8

zetamcfc

19

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?

Reply 9

B_9710

18

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.

Reply 10

Louisb19

14

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

e^{x\ln{6}}

then yes! Now just simplify using the chain rule.

Reply 11

username2097061

OP

2

Original post by Louisb19

If what you mean is

e^{x\ln{6}}

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!

Reply 12

username2097061

OP

2

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

Reply 13

Louisb19

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

e^{\ln{f(x)}} = f(x)

so for f(x) = 6^x gives

e^{\ln{6^x}} = e^{x\ln{6}}

Reply 14

XxKingSniprxX

21

Isn't the answer just 6^xln(6)?

6^x
d/dx = 6^xln(6)

Reply 15

sunburner

1

Original post by atsruser

No.

6^x

and

e^x

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