deriving weirdd thing (1 Viewer)

[enigma_1]

How do you derive this?

It's done weirdly in the solution.

Like especially how do you derive the denominator? Is it xe^x-1 ?

thanks

 

[nerdasdasd]

I think you need to revise your maths . did you use the quotient rule?

 

[bokat]

You need to use the quotient rule which says if y=u/v then y'=(u'v-uv')/v^2
In this case u=x^2 so u'=2x and v=e^x so v'=e^x
So y'=(2x.e^x-x^2.e^x)/(e^x)^2
After factorising for e^x at the denominator and numerator and eliminating it you get:
y'=(2x-x^2)/e^x. We need y' when x=0 so sub 0 for x to get y'=0.
Hope it is useful.
Reza Bokat,
Cambridge Coaching
www.cambridgecoaching.com.au

 

[enigma_1]

You need to use the quotient rule which says if y=u/v then y'=(u'v-uv')/v^2
In this case u=x^2 so u'=2x and v=e^x so v'=e^xSo y'=(2x.e^x-x^2.e^x)/(e^x)^2
After factorising for e^x at the denominator and numerator and eliminating it you get:
y'=(2x-x^2)/e^x. We need y' when x=0 so sub 0 for x to get y'=0.
Hope it is useful.
Reza Bokat,
Cambridge Coaching
www.cambridgecoaching.com.au

Thanks heaps dude. But why didn't you do this, when you derive the v=e^x, why didn't you do v'=xe^x-1 ?

I was taught to do it that way --- helpppppp I have an exam tomorrow

Thanks in advanceee

 

[Carrotsticks]

I think you need to revise your maths . did you use the quotient rule?

I don't think this was necessary at all.

 

[HeroicPandas]

Thanks heaps dude. But why didn't you do this, when you derive the v=e^x, why didn't you do v'=xe^x-1 ?

I was taught to do it that way --- helpppppp I have an exam tomorrow

Thanks in advanceee

Whoever taught you that is working for the DEVIL! That rule only works when the power is a constant (in this case, 'x' is a variable). When u differentiate e^x, the result is just e^x (this is the easiest thing u need to differentiate, its just itself!!!!)

Proof:
y = e^x

ln(y) = x

x = ln(y)

Now differentiate x in terms of y

dx/dy = 1/y

dy/dx = y (but y = e^x)

dy/dx = e^x

So here is the general rule:

y = e^[f(x)]

dy/dx = f'(x) e^[f(x)]

Or in words: differentiate the power and bring it to the front, done

 

[bokat]

Thanks heaps dude. But why didn't you do this, when you derive the v=e^x, why didn't you do v'=xe^x-1 ?

I was taught to do it that way --- helpppppp I have an exam tomorrow

Thanks in advanceee

e^x is the derivative of itself. In other words when you differentiate e^x, it does not change.

 

[enigma_1]

Ohhhhh I think you learn this is integration or something? Because I haven't done that chapter yet. Sorry about that. I just saw it in past papers and thought it was part of calculus or whatever. Thanks everyone, really appreciate it

 

[RealiseNothing]

Ohhhhh I think you learn this is integration or something? Because I haven't done that chapter yet. Sorry about that. I just saw it in past papers and thought it was part of calculus or whatever. Thanks everyone, really appreciate it

It is a part of calculus, it's just not taught straight away during the differentiation chapter.

 

[enigma_1]

Lol ok that explains it