Question

find Dy/DX at y=xcosx​

Answer 1

Given ,

The function is y = x.Cos(x)

Differentiating with respect to x , we get

$</p><p>\sf \Rightarrow \frac{dy}{dx} = \frac{d \{ x.Cos(x)\}}{dx} \\ \\ </p><p>\sf \Rightarrow \frac{dy}{dx} = Cos(x) \frac{dx}{dx} + x \frac{d \{Cos(x) \}}{dx} \: \: \: \{ \because Product \: rule \} \\ \\ </p><p>\sf \Rightarrow \frac{dy}{dx} = Cos(x) - x.Sin(x)$

Remmember :

$\sf \star \: \: \frac{d(u.v)}{dx} = v \frac{d(u)}{dx} + u\frac{d(v) }{dx} \\ \\</p><p></p><p>\sf \star \: \: \frac{d \{Cos(x) \}}{dx} = - Sin(x) \\ \\ \star \: \: \sf \frac{dx}{dx} = 1$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{dy}{dx}=-x\:sin\:x+cos\:x}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies y = x \: cos \: x \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \frac{dy}{dx} =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \frac{d}{dy} (u.v) =v \times \frac{d}{dy}(u) + u \times \frac{d}{dy} (v) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies y' = cos \: x \times \frac{d}{dy} (x) + x \times \frac{d}{dy} (cos \: x) \\ \\ \tt \circ \: Differentiation \: of \: x = 1 \\ \\ \tt: \implies y' = cos \: x + x \times \frac{d}{dy} (cos \: x) \\ \\ \tt \circ \: Differentiation \: of \: cos \: x= - sin \: x \\ \\ \tt: \implies y' = cos \: x + x \times ( - sin \: x) \\ \\ \tt: \implies y' = cos \: x- x \: sin \: x \\ \\ \green{\tt: \implies \frac{dy}{dx} = - x \: sin \: x + cos \: x}$