Question

Find du/dx
when u = sin (x2 + y2), where a²x² + b²y² = c²​

Answer 1

Step-by-step explanation:

We have,

$u = \sin( {x}^{2} + {y}^{2} ) \: \: and \: \: {a}^{2} {x}^{2} + {b}^{2} {y}^{2} = {c}^{2}$

$\implies {y}^{2} = \frac{ {c}^{2} - {a}^{2} {x}^{2} }{ {b}^{2} }$

$\implies2y \frac{dy}{dx} = - 2 \frac{ {a}^{2} }{ {b}^{2} } x$

$\implies \frac{dy}{dx} = \frac{ {a}^{2} x}{ {b}^{2}y }$

Also,

$\frac{du}{dx} = \cos( {x}^{2} + {y}^{2} ) . \{2x + 2y \frac{dy}{dx} \}$

$\implies \frac{du}{dx} = \cos( {x}^{2} + {y}^{2} ) . \{ 2x + 2y. \frac{ {a}^{2}x }{ {b}^{2}y } \}$

$\implies \frac{du}{dx} = 2 x\cos( {x}^{2} + {y}^{2} ) . \{ \frac{ {b}^{2} + {a}^{2} }{ {b}^{2} } \}$

Answer 2

Answer:

Step-by-step explanation: