Question

differentiate this with respect to x sec(x+y)=xy

Answer 1

ANSWER :–

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{ \sec(x + y). \tan(x + y) - y }{x -\sec(x + y). \tan(x + y)}$

EXPLANATION :–

GIVEN :–

• A function sec(x + y) = xy

TO FIND :–

• dy/dx = ?

SOLUTION :–

$\\ \bf \implies \sec(x + y) = xy$

• Using property –

$\\ \bf \: \: \blacktriangleright \: \: \dfrac{d(u.v)}{dx} =u \dfrac{dv}{dx} + v \dfrac{du}{dx}$

$\\ \bf \: \: \blacktriangleright \: \: \dfrac{d( \sec x)}{dx} = \sec(x). \tan(x)$

• Now Differentiate with respect to 'x' –

$\\ \bf \implies \sec(x + y). \tan(x + y) \left[1 + \dfrac{dy}{dx} \right] = x \dfrac{dy}{dx} + y$

$\\ \bf \implies \sec(x + y). \tan(x + y) +\sec(x + y). \tan(x + y) \dfrac{dy}{dx} = x \dfrac{dy}{dx} + y$

$\\ \bf \implies \sec(x + y). \tan(x + y) - y = x \dfrac{dy}{dx} -\sec(x + y). \tan(x + y) \dfrac{dy}{dx}$

$\\ \bf \implies \sec(x + y). \tan(x + y) - y = \dfrac{dy}{dx} \{x -\sec(x + y). \tan(x + y) \}$

$\\ \implies{ \boxed{ \bf \dfrac{dy}{dx} = \dfrac{ \sec(x + y). \tan(x + y) - y }{x -\sec(x + y). \tan(x + y)}}}$

$\\ \rule{220}{2}$