Differentiation of dF÷dx(qQ-q^2) and the answer should be 0.5 i.e ratio of q÷Q

To differentiate the expression \( \frac{dF}{dx}(qQ - q^2) \) and show that the result is \( 0.5 \) when expressed as the ratio \( \frac{q}{Q} \), we need to apply some fundamental rules of calculus, particularly the product rule and the chain rule. Let’s break this down step by step.

Understanding the Expression

The expression \( F = qQ - q^2 \) represents a function of two variables, \( q \) and \( Q \). Here, \( q \) could be a variable dependent on \( x \), while \( Q \) is treated as a constant during differentiation. Our goal is to find the derivative of this function with respect to \( x \).

Applying the Product Rule

To differentiate \( F \), we can use the product rule, which states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

• \( \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \)

In our case, we can treat \( qQ \) as one function and \( -q^2 \) as another. Thus, we can differentiate each part separately.

Calculating the Derivative

Let’s differentiate \( F \):

• For the term \( qQ \):
  • \( \frac{d(qQ)}{dx} = Q \frac{dq}{dx} \) (since \( Q \) is constant)
• For the term \( -q^2 \):
  • \( \frac{d(-q^2)}{dx} = -2q \frac{dq}{dx} \) (using the power rule)

Combining these results, we have:

\( \frac{dF}{dx} = Q \frac{dq}{dx} - 2q \frac{dq}{dx} \)

Factoring out \( \frac{dq}{dx} \), we get:

\( \frac{dF}{dx} = \left( Q - 2q \right) \frac{dq}{dx} \)

Setting the Derivative to Zero

To find the condition under which this derivative equals zero, we set:

\( Q - 2q = 0 \)

Solving for \( q \) gives:

\( Q = 2q \)

From this, we can express the ratio of \( q \) to \( Q \):

\( \frac{q}{Q} = \frac{1}{2} \)

Final Thoughts

This shows that when the derivative \( \frac{dF}{dx} \) is zero, the ratio \( \frac{q}{Q} \) indeed equals \( 0.5 \). This relationship is significant in various applications, particularly in physics and engineering, where such ratios often represent balance points or equilibrium conditions.