common question (1 Viewer)

[Vampire]

I need help on this q:

i. Differentiate with respect to x the function h(x)=(log_10 x)/x

ii. Given that the only stationary point of h(x) is a maximum, deduce e^pi>pi^e

Thx

 

[香港!]

I need help on this q:

i. Differentiate with respect to x the function h(x)=(log_10 x)/x

ii. Given that the only stationary point of h(x) is a maximum, deduce e^pi>pi^e

Thx

h(x)=(log_10 x)\x=(ln x)\x(ln 10)
h'(x)=(1\ln10) [1-lnx]\x²
Set x'=0
(1\ln10) [1-lnx]\x²=0
1=lnx
.: x=e
At x=e, y=h(x)=1\e(ln10)
This y is "maximum"
This y value would be greater than the y value at x=pi
So at x=pi, y=(ln pi)\pi (ln 10)

1\e(ln10)>(ln pi)\pi(ln10)
1\e>(ln pi)\pi
pi\e>ln pi
e^(pi\e)>pi
put each one to the power of e
e^pi>pi^e as required

 

[Stefano]

h(x)=(log_10 x)\x=(ln x)\x(ln 10)
h'(x)=(1\ln10) [1-lnx]\x²

I got h'(x) = (1-log_10 x)/x^2

 

[Slidey]

h(x)=(log_10 x)/x=lnx/(xln10)=y
lny=ln(lnx)-ln(xln10)
y'/y=1/x.1/(lnx)-1/x=(1-lnx)/(xlnx)
y'=(1-lnx)/(xlnx)*lnx/(xln10)
h'(x)=y'=(1/ln10)(1-lnx)/x^2

 

[Stefano]

h(x)=(log_10 x)/x=lnx/(xln10)=y
lny=ln(lnx)-ln(xln10)
y'/y=1/x.1/(lnx)-1/x=(1-lnx)/(xlnx)
y'=(1-lnx)/(xlnx)*lnx/(xln10)
h'(x)=y'=(1/ln10)(1-lnx)/x^2

You lost me at y'/y.

Refresh my memory:
If y=ln[g(x)]
then y'= ?

 

[dawso]

I need help on this q:

i. Differentiate with respect to x the function h(x)=(log_10 x)/x

ii. Given that the only stationary point of h(x) is a maximum, deduce e^pi>pi^e

Thx

part (i) - by "steele's method"

 

[dawso]

You lost me at y'/y.

Refresh my memory:
If y=ln[g(x)]
then y'= ?

y' = g'(x) / g(x)