Applications Of Derivatives

Step 1: For increasing function, we need .

Step 2: Differentiate the function:
Factor the derivative:

Step 3: Solve :
Critical points:
Step 4: Sign chart analysis:

• Interval (-∞, 0): pick
  → Negative
• Interval (0, 1): pick
  → Positive
• Interval (1, 2): pick
  → Negative
• Interval (2, ∞): pick
  → Positive

Conclusion:
for intervals and
Therefore, f(x) is strictly increasing in (0, 1) and (2, ∞)
Similarly, f(x) is decreasing in (−∞, 0) and (1, 2)

Step 1: Understand the function
The given function is f(x) = (log x) / x, where log denotes the natural logarithm

Step 2: Find the first derivative
To find local maxima, we first differentiate f(x) using the quotient rule:
f'(x) = [ (1/x) * x - log x * 1 ] / x² = (1 - log x) / x².

Step 3: Find critical points
Set the derivative equal to zero to find critical points:
(1 - log x) / x² = 0 implies 1 - log x = 0
=> log x = 1
=> x = e (where e ≈ 2.718)

Step 4: Determine the nature of critical point
To check if x = e is a maximum, examine the second derivative or test values around x = e.
For x slightly less than e, f'(x) > 0 (function increasing).
For x slightly greater than e, f'(x) < 0 (function decreasing).
Hence, x = e is a point of local maximum.

Step 5: Conclusion
The function f(x) attains a local maximum at x = e.

Final Answer: x = e

Step 1: Write the formula for volume of a hemisphere.
The volume of a hemisphere of radius is Differentiate with respect to time : Step 2: Use the given condition.
The problem states that the volume increases at a uniform rate. Thus Therefore Step 3: Express .
Thus Step 4: Surface area of a hemisphere.
Surface area of a hemisphere is Differentiate with respect to time: Step 5: Substitute the value of .}
Substitute Thus Step 6: Conclusion.
Hence the rate of change of surface area is inversely proportional to the radius.
Final Result: Thus the surface area varies inversely as the radius.