Solve the given question ↑↑Chapter - Differentiation Class - 11 th

1.	Solve the given question ↑↑Chapter - Differentiation Class - 11 th Thanks ^_^
Answer» $\displaystyle\huge\red{\underline{\underline{Solution}}}$ *FORMULA TO BE IMPLEMENTED* 1. $\displaystyle \: \frac{d}{dx} ( {x}^{n} ) = n \: {x}^{n - 1}$ 2. $\displaystyle \: \frac{d}{dx} ( logx)= \frac{1}{x}$ 3. $\displaystyle \: \frac{d}{dx} (uv \: ) =u \frac{dv}{dx} + v\frac{du}{dx}$ *TO EVALUATE* Differentiate : $y = {x}^{2} \sqrt{x} + {x}^{4} \: logx$ *EVALUATION* LET $u \: = {x}^{2} \sqrt{x} \: \: and \: \: v = {x}^{4} \: logx$ $\implies \: \displaystyle \: \frac{dv}{dx} = 4 {x}^{3} \: logx + {x}^{3}$ So $y = u + v$ Differentiating both sides with respect to x $\displaystyle \: \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ Now $\displaystyle \: u \: = {x}^{2} \sqrt{x} = {x}^{(2 + \frac{1}{2} \: ) } = {x}^{ \frac{5}{2} }$ Differentiating both sides with respect to x we get $\displaystyle \: \frac{du}{dx} = \frac{5}{2} {x}^{( \frac{5}{2} - 1)} = \frac{5}{2} {x}^{ \frac{3}{2} }$ Again $v = {x}^{4} \: logx$ Differentiating both sides with respect to x we get $\displaystyle \: \frac{dv}{dx} = 4 {x}^{3} \: logx + {x}^{4} \times \frac{1}{x}$ *RESULT* $\displaystyle \: \frac{dy}{dx} = \frac{5}{2} {x}^{ \frac{3}{2} } + 4 {x}^{3} \: logx \: + {x}^{3}$