Question

6 Differtiate with resped to x, if
y=2/x^3
+ Sinx​

Answer 1

Solution :

$\sf{Function\:of\:y = \dfrac{2}{x^{3}} + sin(x)}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d}{dx}\bigg[\dfrac{2}{x^{3}} + sin(x)\bigg]}$

$\sf Here, \: the \: function \: of \: y \: can \: be \: written \: as 2x^{-3} \: + sin(x), \\ \sf (As \: according \: to \: the \: rules \: of \: exponent \: i.e, \: \dfrac{1}{x^{n}} = x^{-n})$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d}{dx}[2x^{-3} + sin(x)]}$

$\textsf{Now, by using the sum rule of differentiation, we get :} \bullet \underline{\sf{Power\:rule\:of\: Differentiation}} \\ \\ \sf{\dfrac{d}{dx}[f(x) + g(x)] = \dfrac{d}{dx}[f(x)] + \dfrac{d}{dx}[g(x)]}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(2x^{-3})}{dx} + \dfrac{d[sin(x)]}{dx}}$

$\textsf{Now, by applying the power rule of differentiation and derivative of sin(x), we get :} \\ \\ \bullet \underline{\sf{Power\:rule\:of\: Differentiation}} \\ \sf{\dfrac{d(x^{n})}{dx} = n\cdot x^{(n - 1)}} \\ \\ \bullet\underline{\sf{Derivative\:of\:sin(x)}} \\ \\ \sf{\dfrac{d[sin(x)]}{dx} = cos(x)}$

$:\implies \sf{\dfrac{dy}{dx} = -3\cdot 2x^{(-3 - 1)} + cos(x)}$

$:\implies \sf{\dfrac{dy}{dx} = -6x^{-4} + cos(x)}$

$\boxed{\therefore \sf{\dfrac{d}{dx}\bigg[\dfrac{2}{x^{3}} + sin(x)\bigg] = -6x^{-4} + cos(x)}}$