Explain how you differentiate y=cos(x+y) with respect to x.

Open in App

Solution

$y = \cos (x + y) Differentiating w . r . t . x, we have \frac{dy}{dx} = \frac{d}{dx} \cos (x + y) \frac{dy}{dx} = - \sin (x + y) . \frac{d}{dx} (x + y) \frac{dy}{dx} = - \sin (x + y) . (1 + \frac{dy}{dx}) \frac{dy}{dx} + \frac{dy}{dx} \sin (x + y) = - \sin (x + y) \frac{dy}{dx} = \frac{- \sin (x + y)}{1 + \sin (x + y)}$

Suggest Corrections

Similar questions

y = (sin \frac{x}{2} + cos \frac{x}{2})

\frac{d y}{d x}

x = \frac{π}{3}

{tan}^{- 1} [\frac{cos x + sin x}{cos x - sin x}]

x

x \sqrt{1 + y} + y \sqrt{1 + x} = 0

x

Differentiate with respect to $x$
$y = sec (tan \sqrt{x})$

\frac{sin x + e^{x}}{cos x + e^{x}}

x

Join BYJU'S Learning Program