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differentiate 2^x

how would you differentiate 2^x
Reply 1
2^x = e^xln2 :smile:
Reply 2
Original post by uxa595
how would you differentiate 2^x


Remember differentiating, y=ax y =a^x , gives, dydx=axlna \displaystyle \frac{dy}{dx} = a^x lna
Reply 3
Original post by Coulson72
2^x = e^xln2 :smile:


I got this but i don't think it's right :s-smilie:

Original post by raheem94
Remember differentiating, y=ax y =a^x , gives, dydx=axlna \displaystyle \frac{dy}{dx} = a^x lna


How do you know a^x foes to a^x lna?
Reply 4
Original post by nm786
remeber:dy/dx=xn=nxn1remeber: dy/dx=x^n=nx^n-1
2x=(1)2x(11)=21=22^x=(1)2x^(1-1)=2*1=2


No.
Reply 5
Original post by uxa595
I got this but i don't think it's right :s-smilie:



How do you know a^x foes to a^x lna?


Its a formula.

If y=ax y= a^x
Take the natural log of both sides.
lny=lnax    lny=xlna lny = lna^x \implies lny=xlna
Differentiate implicitly,
1ydydx=lna    dydx=ylna \displaystyle \frac1y \frac{dy}{dx} = lna \implies \frac{dy}{dx} = ylna

We know, y=ax y=a^x , hence, dydx=ylna    dydx=axlna \displaystyle \frac{dy}{dx} = ylna \implies \frac{dy}{dx} = a^x lna

Do you get it?
Reply 6
Original post by uxa595
I got this but i don't think it's right :s-smilie:



How do you know a^x foes to a^x lna?


If y= a^x

ln(y)= xln(a)

1/y(dy/dx)= ln(a)

dy/dx= a^x(ln(a))
Reply 7
Original post by nm786
is this for c1?


no
Reply 8
Original post by f1mad
If y= a^x

ln(y)= xln(a)

1/y(dy/dx)= ln(a)

dy/dx= a^x(ln(a))


My is better :tongue:
Reply 9
I got this:

y=2^x
therefore
y= e^(xln2)
y= e^x . e^ln2
y= 2e^x

is that right?
Original post by nm786
is this for c1?


C1? Holy **** I haven't done this for C4.
Reply 11
Original post by uxa595
I got this:

y=2^x
therefore
y= e^(xln2)
y= e^x . e^ln2
y= 2e^x

is that right?


No,

Example:

y=3x=eln3x=exln3dydx=ln3×exln3=ln3×eln3x=3xln3 \displaystyle y=3^x = e^{ln3^x} = e^{xln3} \\ \frac{dy}{dx} = ln3 \times e^{xln3} = ln3 \times e^{ln3^x} = 3^xln3

This could have directly been found using the formula i gave you.
Reply 12
Original post by raheem94
Its a formula.

If y=ax y= a^x
Take the natural log of both sides.
lny=lnax    lny=xlna lny = lna^x \implies lny=xlna
Differentiate implicitly,
1ydydx=lna    dydx=ylna \displaystyle \frac1y \frac{dy}{dx} = lna \implies \frac{dy}{dx} = ylna

We know, y=ax y=a^x , hence, dydx=ylna    dydx=axlna \displaystyle \frac{dy}{dx} = ylna \implies \frac{dy}{dx} = a^x lna

Do you get it?

Or... (if you're not fond of implicit differentiation):

y=ax    y=elnax=exlna\displaystyle y=a^x \implies y=e^{\ln a^x} = e^{x\ln a}

    dydx=lna exlna=axlna\displaystyle \implies \frac{dy}{dx} = \ln a \ e^{x\ln a} = a^x \ln a
Reply 13
Original post by uxa595
I got this:

y=2^x
therefore
y= e^(xln2)
y= e^x . e^ln2
y= 2e^x

is that right?


No.

If y= e^(xln2)

dy/dx= ln2*e^(xln2)

And you should remember: e^(xln2)= 2^x.
Reply 14
Original post by notnek
Or... (if you're not fond of implicit differentiation):

y=ax    y=elnax=exlna\displaystyle y=a^x \implies y=e^{\ln a^x} = e^{x\ln a}

    dydx=lna exlna=axlna\displaystyle \implies \frac{dy}{dx} = \ln a \ e^{x\ln a} = a^x \ln a


See post#14 in this thread, i know it mate.
Reply 15
Original post by raheem94
No,

Example:

y=3x=eln3x=exln3dydx=ln3×exln3=ln3×eln3x=3xln3 \displaystyle y=3^x = e^{ln3^x} = e^{xln3} \\ \frac{dy}{dx} = ln3 \times e^{xln3} = ln3 \times e^{ln3^x} = 3^xln3

This could have directly been found using the formula i gave you.


Ahhh right, thanks, i've got it now :smile:

I can't remember doing this ever in class :tongue:
Reply 16
Original post by raheem94
See post#14 in this thread, i know it mate.

I assumed you knew it. I posted it for the OP.
Just tried this myself

Doing it implicitly seems the most straight forward

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