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C3 Differentiation

Need help to differentiate this :

3e^x - 1/2lnx - 2

So differentiating the first term will give : 3e^x

How about the second one ? -1/2lnx, how exactly do I differentiate that ?
Reply 1
Avatar for Andylol
Andylol
14
y = 1/2 ln x
dy/dx =
1/2
---
x
Reply 2
Avatar for JeevesTSR
JeevesTSR
uer23


How about the second one ? -1/2lnx, how exactly do I differentiate that ?


Is it -1 / 2lnx or -0.5lnx ?
Reply 3
Avatar for Skadoosh
Skadoosh
12
Use Chain Rule. And from then, by inspection.
differentiate 2lnx separately if that makes it any easier then flip the fraction
Reply 5
Avatar for Dreizhen
Dreizhen
13
EDIT: Disregard this I am an idiot
Reply 6
Avatar for uer23
uer23
OP
15
Andylol
y = 1/2 ln x
dy/dx =
1/2
---
x


Right so it's −12x- \frac{1}{2x}
3e^x - 1/2x

jst differentiate lnx then multiply by -0.5
Reply 8
Avatar for uer23
uer23
OP
15
JeevesTSR
Is it -1 / 2lnx or -0.5lnx ?


-0.5lnx.
Reply 9
Avatar for JeevesTSR
JeevesTSR
uer23
-0.5lnx.


ok then i think its -0.5 * 1/x

= -1 / 2x
uer23
Right so it's −12x- \frac{1}{2x}


correct
Reply 11
Avatar for uer23
uer23
OP
15
So related to this question...

The differential of the curve f(x)=3ex−12ln⁡x−2f(x) = 3e^x - \frac{1}{2} \ln x - 2

is as stated above f′(x)=3ex−12xf'(x) = 3e^x - \frac{1}{2x}

The curve f(x) has a turning point at p, the x co-ordinate of p is α\alpha

Show that α=16e−α\alpha = \frac{1}{6}e^{- \alpha}

So far what I've done

3eα−12α=03e^ \alpha - \frac{1}{2 \alpha} = 0

3eα=12α3e^ \alpha = \frac{1}{2 \alpha}

6αeα=16 \alpha e^ \alpha = 1

eα=16αe^ \alpha = \frac{1}{6 \alpha}

So what do I do know take the natural log of both sides ? But then how is that going to get me closer to my target which I have to show.
Reply 12
Avatar for uer23
uer23
OP
15
Bump... +rep available.
Reply 13
Avatar for Toneh
Toneh
8
you're basically there....just multiply by alpha, then divide by e^alpha and you've done it...
Reply 14
Avatar for Andylol
Andylol
14
uer23
So related to this question...

The differential of the curve f(x)=3ex−12ln⁡x−2f(x) = 3e^x - \frac{1}{2} \ln x - 2

is as stated above f′(x)=3ex−12xf'(x) = 3e^x - \frac{1}{2x}

The curve f(x) has a turning point at p, the x co-ordinate of p is α\alpha

Show that α=16e−α\alpha = \frac{1}{6}e^{- \alpha}

So far what I've done

3eα−12α=03e^ \alpha - \frac{1}{2 \alpha} = 0

3eα=12α3e^ \alpha = \frac{1}{2 \alpha}

6αeα=16 \alpha e^ \alpha = 1

eα=16αe^ \alpha = \frac{1}{6 \alpha}

So what do I do know take the natural log of both sides ? But then how is that going to get me closer to my target which I have to show.


All the calculations are done, just need to rearrange alpha and e^a

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