Question

differentiate with respect to x,
y=√(x+1)(x-2)(x+3)​

Answer 1

Given ,

The function is

$y = \sqrt{(x + 1)(x + 2)(x + 3)}$

Differentiating y wrt x , we get

$\tt \implies \frac{dy}{dx} = \frac{d \{\sqrt{(x + 1)(x + 2)(x + 3) } \: \} }{dx}$

$\tt \implies\frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \{(x + 1)(x + 2)(x + 3) \}}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \bigg\{ \{{(x)}^{2} + 2x + x + 2 \}(x + 3) \bigg\}}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \bigg\{ {(x)}^{3} +2 {(x)}^{2} + {(x)}^{2} + 2x + 3 {(x)}^{2} + 6x + 3x + 6 \ \bigg\}}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \bigg\{ {(x)}^{3} +6{(x)}^{2} + 11x + 6 \bigg\}}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \{ \frac{d {(x)}^{3} }{dx} + \frac{d \{6{(x)}^{2} \}}{dx} + \frac{d(11x)}{dx} + \frac{d(6)}{dx} \}$

$\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \{ 3 {(x)}^{2} + 12x + 11 \}$

$\tt \implies \frac{dy}{dx} = \frac{ 3 {(x)}^{2} + 12x + 11}{2 \sqrt{(x + 1)(x + 2)(x + 3)} }$

Remmember :

$\tt \frac{d {(x)}^{n} }{dx} = n {(x)}^{n - 1}$

$\tt \frac{d( \sqrt{x} )}{dx} = \frac{1}{2 \sqrt{x} }$

$\tt \frac{d(u \pm v}{dx} = \frac{d(u)}{dx} \pm \frac{d(v)}{dx}$

$\tt\frac{d(constant)}{dx} = 0$

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Answer 2

Given ,

The function is

$y = \sqrt{(x + 1)(x + 2)(x + 3)}$

Differentiating y wrt x , we get

$\tt \implies \frac{dy}{dx} = \frac{d \{\sqrt{(x + 1)(x + 2)(x + 3) } \: \} }{dx}</p><p></p><p> \\ </p><p> </p><p></p><p>\tt \implies\frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \{(x + 1)(x + 2)(x + 3) \}}{dx}</p><p> </p><p> </p><p></p><p> \\ </p><p> </p><p></p><p>\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \bigg\{ \{{(x)}^{2} + 2x + x + 2 \}(x + 3) \bigg\}}{dx}</p><p> </p><p></p><p> \\ </p><p></p><p>\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \bigg\{ {(x)}^{3} +2 {(x)}^{2} + {(x)}^{2} + 2x + 3 {(x)}^{2} + 6x + 3x + 6 \ \bigg\}}{dx}</p><p> </p><p> \\ </p><p></p><p>\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \frac{d \bigg\{ {(x)}^{3} +6{(x)}^{2} + 11x + 6 \bigg\}}{dx}</p><p> </p><p> \\ </p><p></p><p>\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \{ \frac{d {(x)}^{3} }{dx} + \frac{d \{6{(x)}^{2} \}}{dx} + \frac{d(11x)}{dx} + \frac{d(6)}{dx} \} \\ </p><p></p><p>\tt \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{(x + 1)(x + 2)(x + 3)} } \{ 3 {(x)}^{2} + 12x + 11 \} \\ </p><p>\tt \implies \frac{dy}{dx} = \frac{ 3 {(x)}^{2} + 12x + 11}{2 \sqrt{(x + 1)(x + 2)(x + 3)} }$

Remmember :

$\tt \frac{d {(x)}^{n} }{dx} = n {(x)}^{n - 1} </p><p></p><p> \\ </p><p></p><p>\tt \frac{d( \sqrt{x} )}{dx} = \frac{1}{2 \sqrt{x} } </p><p></p><p> \\ </p><p></p><p>\tt \frac{d(u \pm v}{dx} = \frac{d(u)}{dx} \pm \frac{d(v)}{dx} </p><p></p><p> \\ </p><p> </p><p></p><p>\tt\frac{d(constant)}{dx} = 0$

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