Applications of Derivatives
Implicit Differentiation
Grade 12

Question:

<p>If <span>\(2^x + 2^y = 2^{x+y}\)</span>, then <span>\(\frac{dy}{dx}\)</span> is equal to</p>
<p>(a) <span>\(\frac{y}{x}\)</span></p>
<p>(b) <span>\(\frac{x}{y}\)</span></p>
<p>(c) <span>\(\frac{1 - 2^y}{2^x - 1}\)</span></p>
<p>(d) <span>\(\frac{2y(1 - x)}{2}\)</span></p>

Step-by-Step Solution

Key Concept: Use implicit differentiation on an equation with exponential terms.
<p>Given: <span>$2^x + 2^y = 2^{x+y}$</span></p><p>Differentiating both sides with respect to <span>$x$</span>:</p><p><span>$2^x\ln 2 + 2^y\ln 2 \cdot \frac{dy}{dx} = 2^{x+y}\ln 2\left(1 + \frac{dy}{dx}\right)$</span></p><p><span>$2^x + 2^y\frac{dy}{dx} = 2^{x+y}\left(1 + \frac{dy}{dx}\right)$</span></p><p><span>$2^x + 2^y\frac{dy}{dx} = 2^{x+y} + 2^{x+y}\frac{dy}{dx}$</span></p><p><span>$2^y\frac{dy}{dx} - 2^{x+y}\frac{dy}{dx} = 2^{x+y} - 2^x$</span></p><p><span>$\frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y - 2^{x+y}} = \frac{1 - 2^y}{2^x - 1}$</span></p>
Correct Answer: C

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