Limits, Continuity & Differentiability
Implicit Differentiation
Grade 12
Question:
<p>If <i>y</i> is a function of <i>x</i> and <i>log</i>(<i>x</i> + <i>y</i>) = 2<i>xy</i>, then the value of <i>y</i>'(0) is</p>
<p>(a) 1</p>
<p>(b) 1/3</p>
<p>(c) 2</p>
<p>(d) 0</p>
Step-by-Step Solution
Key Concept: Differentiate the given implicit relation and substitute the initial condition x=0 to find y'(0).
<p>Differentiate both sides with respect to <i>x</i>:</p><p>$$\frac{1}{x+y}\left(1 + \frac{dy}{dx}\right) = 2y + 2x\frac{dy}{dx}$$</p><p>At <i>x</i> = 0: From the original equation, log(<i>y</i>) = 0, so <i>y</i> = 1.</p><p>Substituting: $$\frac{1}{1}\left(1 + y'(0)\right) = 2(1) + 0$$</p><p>$$1 + y'(0) = 2$$</p><p>$$y'(0) = 1$$</p>
Correct Answer: A