If $(\cos x)^{y}=(\sin y)^{x}$, find $\dfrac{dy}{dx}$.

$\dfrac{\log(\sin y)+y\tan x}{\log(\cos x)-x\cot y}$

CBSE Class 12 · 2019 · DifferentiationLong Answer

(cos x)^{y} = (sin y)^{x}

, find

\frac{d y}{d x}

Answer: $\frac{lo g ( sin y ) + y tan x}{lo g ( cos x ) - x cot y}$

Take logarithm:

y lo g (cos x) = x lo g (sin y)

Differentiate both sides (product rule each side):

lo g (cos x) \frac{d y}{d x} - y tan x = lo g (sin y) + x cot y \frac{d y}{d x}

Collect

\frac{d y}{d x}

\frac{d y}{d x} [lo g (cos x) - x cot y] = lo g (sin y) + y tan x

\frac{d y}{d x} = \frac{lo g ( sin y ) + y tan x}{lo g ( cos x ) - x cot y}

CBSE Class 12 Board · 2019 · Subjective