Limits, Continuity & Differentiability
Product Rule of Differentiation
Grade 12
<p>Let <span class="math">\(f\)</span>, <span class="math">\(g\)</span> and <span class="math">\(h\)</span> be differentiable functions. If <span class="math">\(f(0) = 1\)</span>, <span class="math">\(g(0) = 2\)</span>, <span class="math">\(h(0) = 3\)</span> and the derivative of their pairwise product at <span class="math">\(x = 0\)</span> are <span class="math">\((fg)'(0) = 6\)</span>, <span class="math">\((gh)'(0) = 4\)</span> and <span class="math">\((hf)'(0) = 5\)</span>, then compute the value of <span class="math">\((fgh)'(0)\)</span>.</p>
Step-by-Step Solution
Key Concept: Use the product rule for three functions: the derivative of a product of three functions equals the sum of each pairwise product's derivative multiplied by the third function's value.
The product rule for three differentiable functions $f$, $g$, and $h$ is given by:
$$ (fgh)'(x) = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) $$
At $x=0$, this formula becomes:
$$ (fgh)'(0) = f'(0)g(0)h(0) + f(0)g'(0)h(0) + f(0)g(0)h'(0) $$
To compute $(fgh)'(0)$, we first need to determine the values of $f'(0)$, $g'(0)$, and $h'(0)$. We use the given information about the derivatives of pairwise products and the product rule for two functions, $(uv)'(x) = u'(x)v(x) + u(x)v'(x)$.
Given values at $x=0$:
$f(0) = 1$, $g(0) = 2$, $h(0) = 3$
$(fg)'(0) = 6$
$(gh)'(0) = 4$
$(hf)'(0) = 5$
Applying the product rule for two functions at $x=0$:
1. For $(fg)'(0)$:
$f'(0)g(0) + f(0)g'(0) = 6$
$f'(0)(2) + (1)g'(0) = 6$
$$ 2f'(0) + g'(0) = 6 \quad (1) $$
2. For $(gh)'(0)$:
$g'(0)h(0) + g(0)h'(0) = 4$
$g'(0)(3) + (2)h'(0) = 4$
$$ 3g'(0) + 2h'(0) = 4 \quad (2) $$
3. For $(hf)'(0)$:
$h'(0)f(0) + h(0)f'(0) = 5$
$h'(0)(1) + (3)f'(0) = 5$
$$ 3f'(0) + h'(0) = 5 \quad (3) $$
Now we solve the system of linear equations (1), (2), and (3) for $f'(0)$, $g'(0)$, and $h'(0)$.
From equation (1), we express $g'(0)$ in terms of $f'(0)$:
$g'(0) = 6 - 2f'(0)$
Substitute this expression for $g'(0)$ into equation (2):
$3(6 - 2f'(0)) + 2h'(0) = 4$
$18 - 6f'(0) + 2h'(0) = 4$
$2h'(0) - 6f'(0) = -14$
$$ h'(0) - 3f'(0) = -7 \quad (4) $$
Now we have a system of two equations with $f'(0)$ and $h'(0)$ from (3) and (4):
$$ 3f'(0) + h'(0) = 5 \quad (3) $$
$$ -3f'(0) + h'(0) = -7 \quad (4) $$
Adding equation (3) and equation (4):
$(3f'(0) + h'(0)) + (-3f'(0) + h'(0)) = 5 + (-7)$
$2h'(0) = -2$
$h'(0) = -1$
Substitute $h'(0) = -1$ into equation (3):
$3f'(0) + (-1) = 5$
$3f'(0) = 6$
$f'(0) = 2$
Substitute $f'(0) = 2$ into the expression for $g'(0)$:
$g'(0) = 6 - 2(2)$
$g'(0) = 6 - 4$
$g'(0) = 2$
Thus, we have found the derivatives at $x=0$:
$f'(0) = 2$
$g'(0) = 2$
$h'(0) = -1$
Finally, substitute these values along with $f(0)=1$, $g(0)=2$, and $h(0)=3$ into the product rule for three functions at $x=0$:
$$ (fgh)'(0) = f'(0)g(0)h(0) + f(0)g'(0)h(0) + f(0)g(0)h'(0) $$
$$ (fgh)'(0) = (2)(2)(3) + (1)(2)(3) + (1)(2)(-1) $$
$$ (fgh)'(0) = 12 + 6 - 2 $$
$$ (fgh)'(0) = 18 - 2 $$
$$ (fgh)'(0) = 16 $$
Correct Answer: 16