Question

differentiate y=x³(x²-2x)with respect to x (by using product rule)​

Answer 1

$\sf \rightarrow \quad y = {x}^{3} ( {x}^{2} - 2x) \\ \\ \sf \rightarrow \quad y' = \frac{d( {x}^{3} \cdot ( {x}^{2} - 2x)) }{dx} \\ \\ \sf \: \: Applying \: Product \: Rule \: of \: Differentiation \\ \\ \boxed{\sf \quad \frac{d( \red{ {x}^{n} }\cdot \green{x}) }{dx} = \green{x }\cdot \frac{d( \red{ {x}^{n} })}{dx} + \red{{x}^{n} }\cdot \frac{d( \green{ x}) }{dx} }</p><p> \\ \\ \sf \rightarrow \quad y' = \bigg \{ ( {x}^{2} - 2x) \cdot \frac{d( {x}^{3} )}{dx} \bigg \} + \bigg \{ {x}^{3} \cdot \frac{d( {x}^{2} - 2x) }{dx} \bigg \} \\ \\ \sf \rightarrow \quad y' = ( {x}^{2} - 2x)(3 {x}^{2} ) + {x}^{3} \bigg( \frac{d( {x}^{2}) }{dx} - 2 \frac{d(x)}{dx} \bigg) \\ \\ \sf \rightarrow \quad y' = 3 {x}^{4} - 6 {x}^{3} + {x}^{3} (2x - 2) \\ \\ \sf \rightarrow \quad y' = 3 {x}^{4} - 6 {x}^{3} + 2 {x}^{4} - 2 {x}^{3} \\ \\ \sf \rightarrow \quad \boxed{ \blue{ \sf \pink{y'} = 5 {x}^{4} - 8 {x}^{3}} } \\ \\ \\ \underline{ \sf Properties \: Used} \\ \\ \sf \quad \hookrightarrow \quad \frac{d( \red{ {x}^{ \green{n}} })}{dx} = \green{n} \cdot \red{ {x}}^{ \green{n} - 1} \\ \\ \sf \quad \hookrightarrow \quad \frac{d( \blue{ a} \red{ {x}^{n} }) }{dx} = \blue{a} \cdot \frac{d( \red{ {x}^{n} })}{dx} \\ \\ \qquad \quad \sf \blue{a} = constant$