Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $u(x)$ and $v(x)$ be differentiable functions such that $\dfrac{u(x)}{v(x)} = 7$. If $\dfrac{u'(x)}{v'(x)} = p$ and $\left(\dfrac{u(x)}{v(x)}\right)' = q$, then $\dfrac{p+q}{p-q}$ has the value equal to:</p>
<p>1</p>
<p>0</p>
<p>7</p>
<p>-7</p>
Step-by-Step Solution
Key Concept: General
Given that $u(x)$ and $v(x)$ are differentiable functions such that $\dfrac{u(x)}{v(x)} = 7$.
Step 1: Determine the value of $p$.
Since $\dfrac{u(x)}{v(x)} = 7$, it follows that $u(x) = 7v(x)$.
Differentiating both sides with respect to $x$, we get $u'(x) = 7v'(x)$.
Therefore, $p = \dfrac{u'(x)}{v'(x)} = \dfrac{7v'(x)}{v'(x)} = 7$.
Step 2: Determine the value of $q$.
Given $q = \left(\dfrac{u(x)}{v(x)}\right)'$.
Since $\dfrac{u(x)}{v(x)} = 7$ (a constant), its derivative is $0$.
Thus, $q = 0$.
Step 3: Calculate the value of $\dfrac{p+q}{p-q}$.
Substitute the values of $p=7$ and $q=0$:
$$ \frac{p+q}{p-q} = \frac{7+0}{7-0} = \frac{7}{7} = 1 $$
Correct Answer: D