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Chain Rule of Differentiation

Differentiate...

Question

Differentiate:
${sin}^{2} y + cos x y = K$

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Solution

$\frac{d y}{d x} o f {s i n}^{2} y + C o s x y = π$

$2 S i n x (C o s x y \frac{d y}{d x}) - S i n x y (y + \frac{x d y}{d x}) = 0$

$2 S i n y C o s y \frac{d y}{d x} = S i n x y (y + x \frac{d y}{d x})$

${S i n}^{2} y \frac{d y}{d x} = y S i n x y + x S i n x y \frac{d y}{d x}$

Hence ,

$\frac{d y}{d x} = \frac{y S i n x y}{S i n 2 y} - x S i n x y$

The same value of $\frac{d y}{d x S i n y}$ and $\frac{d y}{d x C o s y}$ will be obtained if using logs on both the sides .

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