I got problem with the derivative of y=a^2x (2 Viewers)

[red802]

can someone show me step to step of the this question please and not just put the answer.

Differentiate y = a^2x

 

[darkliight]

The most familiar way is to probably just use the chain rule.
Let u = a^x, then y = u^2.

We have dy/du = 2u and du/dx = ln(a)*a^x.

dy/dx = dy/du * du/dx = 2u*ln(a)*a^x = 2(a^x)*ln(a)*a^x = 2*ln(a)*a^(2x).

Try a^(3x) and you'll notice a pattern though, which will be handy for exams.

 

[Mountain.Dew]

u might also like to consider chaning a^2x into e^(bx), so its easier to differentiate.

e^(b) = a^(2x)

log both sides...==> ln[e^(b)] = ln[a^(2x)'

using our log laws...that ln(a^b) = b*lna...

b [ln e] = 2x ln a, so b = 2xlna

so a^(2x) = e^2lnax, noting that 2lna is a constant.

so, d[a^(2x)]/dx = d[e^2lnax]/dx = [2lna][e^2lnax] = [2lna][a^(2x)]

 

[red802]

i kinda get it but can u also help me with this question (6x^2+1).a^x

 

[Riviet]

Use the product rule:

let u=6x²+1 => u'=12x

let v=a^x => v'=a^x.log_ea

d/dx(uv)=uv' + vu'

 

[red802]

im confused about the a^x/lna= v'. how does that work out

 

[Riviet]

Sorry, I meant to multiply a^x and ln a. Fixed now.

 

[red802]

kool. i get it now, thanks

 

[abcd9146]

lol, this may sound stupid... but why is the derivative of a^x = a^x.lna???? i forgot if i ever learnt this (i probably have and forgot )


also, another question, what is the derivative of:
y=e^x2

 

[abcd9146]

Iruka, thanks, i dont think i ever learnt that...

so from doing that, is d/dx[a^3x ]
3a^2x.lna

actually, i remembered teh question wrong...
its meant to me

d/dx [x^ex]

 

[Riviet]

x^ex=e^log_exex
=e^ex.log_ex

Now you can use the chain rule [as well as the product rule for the derivative of the index] to differentiate it.

 

[abcd9146]

thanks, i would've never got that...
how'd you know to add the e^log_e? just practice lots?

 

[darkliight]

Iruka, thanks, i dont think i ever learnt that...

so from doing that, is d/dx[a^3x ]
3a^2x.lna

Close, write out your working.

 

[SoulSearcher]

thanks, i would've never got that...
how'd you know to add the eloge? just practice lots?

That's what you should do for maths anyway, however this type of question rarely turns up in 2 unit exams.

 

[abcd9146]

Close, write out your working.

y=a^3x

let u=a^x
y=u³

du/dx=a^x.lna
dy/du=3u²

du/dx.dy/du=dy/dx

dy/dx=3u².a^x.lna
=3(a^x)².a^x.lna
=3a^3x.lna

thats right now... i think... my dy/du was wrong before

 

[darkliight]

Nice And you should be able to see a general rule for d/dx a^(nx) now.

y=a^(-4x)
and
y=a^(pi*x)

should be easy now without even thinking about them.

 

[Riviet]

thanks, i would've never got that...
how'd you know to add the e^log_e? just practice lots?

I knew you can't differentiate x^f(x), so the only way I could have done it was raise it to the power e, so that I could.
But of course, it all comes with practice and experience.

 

[Riviet]

Yep, I've seen it, but the problem is we were given an expression to differentiate, so we had very little choice but to change it into base e. Anyway, implicit differentiation and I by parts is cool.

 

[abcd9146]

logarithmic differentiation... how does that work?? from what you did, you log both sides, then differentiate then seperately, but why do you times by dy/dx on the LHS?

oh and would they ask these types of questions in the 2 or 3 unit HSC

 

[SoulSearcher]

They would probably ask simple differentiations of a^f(x), nothing too difficult if they ask it, because it rarely appears in the HSC exams.

Not too sure why the dy/dx is there, haven't learnt implicit differentiation myself yet.