Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $f$ be defined for all $x \in \mathbb{R}$. If $f$ is differentiable and $f(x^3) = x^5$ for all $x \in \mathbb{R}\,(x\ne0)$, then $f'(27)$ is equal to:</p>
<p>15</p>
<p>45</p>
<p>0</p>
<p>35</p>
Step-by-Step Solution
Key Concept: General
<b>Chain Rule on Composite Function</b><br>Given $f(x^3) = x^5$. Differentiate both sides w.r.t. $x$:<br>$f'(x^3) \cdot 3x^2 = 5x^4$<br>$f'(x^3) = \dfrac{5x^4}{3x^2} = \dfrac{5x^2}{3}$<br>At $x^3 = 27 \Rightarrow x = 3$:<br>$f'(27) = \dfrac{5 \cdot 9}{3} = 15$<br><b>Key concept:</b> Differentiate the composite $f(g(x))$ using chain rule, then substitute the target value.<br><b>Trap:</b> Students plug in $x=27$ directly without finding the correct $x$ from $x^3=27$.
Correct Answer: A