Question

Find: dy/dx, if y=x sinx + cos x

Answer 1

Answer:

dy/ dx = (-x cosx + sinx ) - sinx

dy/dx = -x cosx

Step-by-step explanation:

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Answer 2

Solution:

Given That:

$\rm \longrightarrow y = x \sin(x) + \cos(x)$

Differentiating wrt x, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx} \{x \sin(x) + \cos(x) \}$

We know that:

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(f \pm g) = \dfrac{d}{dx}f \pm \dfrac{d}{dx} g }}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(f \cdot g) = g\dfrac{d}{dx}f + f\dfrac{d}{dx} g }}$

Using this results, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx} x \sin(x) + \dfrac{d}{dx} \cos(x)$

$\rm \longrightarrow \dfrac{dy}{dx} = \bigg[\dfrac{d}{dx} x \sin(x) \bigg]+ \dfrac{d}{dx} \cos(x)$

$\rm \longrightarrow \dfrac{dy}{dx} = \bigg[x\dfrac{d}{dx}\sin(x) + \sin(x) \dfrac{d}{dx}x \bigg]+ \dfrac{d}{dx} \cos(x)$

$\rm \longrightarrow \dfrac{dy}{dx} = \bigg[x\cdot \cos(x) + \sin(x) \cdot 1\bigg] - \sin(x)$

$\rm \longrightarrow \dfrac{dy}{dx} =x\cos(x)$

★ Which is our required answer.

Learn More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$