Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>If $(\cos x)^y = (\sin y)^x$, then $\dfrac{dy}{dx}$ equals:</p>
<p>$\dfrac{\log\sin y + y\tan x}{\log\cos x - x\cot y}$</p>
<p>$\dfrac{\log\sin y - y\tan x}{\log\cos x + x\cot y}$</p>
<p>$\dfrac{\log\cos y + y\tan x}{\log\sin x - x\cot y}$</p>
<p>$\dfrac{\log\sin y + y\tan x}{\log\cos x + x\cot y}$</p>
Step-by-Step Solution
Key Concept: General
<b>Logarithmic Differentiation</b><br>Take $\ln$ both sides: $y\ln(\cos x) = x\ln(\sin y)$<br>Differentiate: $\frac{dy}{dx}\ln(\cos x) + y\cdot\frac{-\sin x}{\cos x} = \ln(\sin y) + x\cdot\frac{\cos y}{\sin y}\cdot\frac{dy}{dx}$<br>$\frac{dy}{dx}\left[\ln(\cos x) - x\cot y\right] = \ln(\sin y) + y\tan x$<br>$\frac{dy}{dx} = \dfrac{\ln(\sin y) + y\tan x}{\ln(\cos x) - x\cot y}$<br><b>Key concept:</b> $\ln$ both sides first, then implicit differentiation.<br><b>Trap:</b> Students confuse $\sin$ and $\cos$ arguments when differentiating the RHS.
Correct Answer: B