Exponential and Logarithmic Functions (1 Viewer)

[Wolf-]

Hey guys, I've been doing some maths and have been stuck on these 3 questions.

1
If f(x) = ln(2-x)^1/2 find f'(1)

2.
Differentiate e^2x / lnx

3.
Find the stationary point on the curve y = lnx / x and determine it's nature

Any help on these is greatly appreciated.

Cheers

 

[lyounamu]

Hey guys, I've been doing some maths and have been stuck on these 3 questions.

1
If f(x) = ln(2-x)^1/2 find f'(1)

2.
Differentiate e^2x / lnx

3.
Find the stationary point on the curve y = lnx / x and determine it's nature

Any help on these is greatly appreciated.

Cheers

1.

f(x) = ln(2-x)^1/2
= 1/2 ln(2-x)
f'(x) = 1/2 . -1/(2-x) = -1/2(2-x)
if x=1, f'(1) = -1/2

2.
y = e^2x/lnx
dy/dx = (lnx . 2e^2x - e^2x . 1/x )/ 2lnx

3. y = lnx/x
dy/dx = (x . 1/x - lnx )/x^2 = (1-lnx)/x^2

so if dy/dx = 0, 1-lnx = 0
1 = lnx
x = e

using first derivative,
whe x>e, dy/dx = -
when x<3, dy/dx = +

max point.

 

[Wolf-]

1.

f(x) = ln(2-x)^1/2
= 1/2 ln(2-x)
f'(x) = 1/2 . -1/(2-x) = -1/2(2-x)
if x=1, f'(1) = -1/2

2.
y = e^2x/lnx
dy/dx = (lnx . 2e^2x - e^2x . 1/x )/ 2lnx

3. y = lnx/x
dy/dx = (x . 1/x - lnx )/x^2 = (1-lnx)/x^2

so if dy/dx = 0, 1-lnx = 0
1 = lnx
x = e

using first derivative,
whe x>e, dy/dx = -
when x<3, dy/dx = +

max point.

Thanks alot, that's the same answer I got for no.2, rechecked answers and realised I was looking at the wrong exercise haha