Question

find the second derivative of y= x/square root of x-1​

Answer 1

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$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: y = \frac{ {x}^{2} }{ \sqrt{x - 1} } \\ First \:\:derivatives\\ \sf \: \implies \: \frac{dy}{dx} = \frac{d \: \: ( \frac{ {x}^{2} }{ \sqrt{x - 1} }) }{dx} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ \sqrt{x - 1} \times \frac{d \: {x}^{2} }{dx} \: \: - \: \: {x}^{2} \times \frac{d \: \: \sqrt{x - 1} }{dx} }{ ({ \sqrt{x - 1} })^{2} } \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2x \sqrt{x - 1} \: \: - \: \: {x}^{2} \times \frac{1}{2 \sqrt{x - 1} } }{x - 1} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2x}{ \sqrt{x - 1} } - \frac{ {x}^{2} }{2 {(x - 1)}^{ \frac{3}{2} } } \\ \sf \: \: \: \: \: \: \: \: \therefore \: \frac{dy}{dx} = \frac{x(3x - 4)}{2 {(x - 1)}^{ \frac{3}{2} } } \\ \\ second \: derivatives \\ \\ \sf \: \frac{ {d}^{2} y }{d {x}^{2} } = \frac{d \: \: \{\frac{x(3x - 4)}{2 {(x - 1)}^{ \frac{3}{2} } } \}}{dx} \\ \\ \sf \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times \frac{d \: \{ \frac{x(3x - 4)}{ {(x - 1)}^{ \frac{3}{2} } } \}}{dx} \\ \\ \sf \: \: \: \: \: \: \: \: \: = \frac{ \frac{ {(x - 1)}^{ \frac{3}{2} } \times \{ \frac{d \: x(3x - 4)}{dx} - x(3x - 4) \frac{d {(x - 1)}^{ \frac{3}{2} } }{dx} \}}{( { {(x - 1)}^{ \frac{3}{2} } })^{2} } }{2} \\ \sf \: \: \: \: \: \: \: \: \: = \frac{ {x - 1)}^{ \frac{3}{2} } \times \{ (3x - 4)\frac{d \: x}{dx} + x \frac{d \: (3x - 4)}{dx} \} - x(3x - 4) \frac{3}{2} (x - 1)^{ \frac{1}{2} } \times \frac{d(x - 1)}{dx} }{2 {(x - 1)}^{3} } \\ \sf \: \: \: \: \: \: \: \: \: = \frac{ {(x - 1)}^{ \frac{3}{2} } \times \{x(3 + 0) + 3x - 4 \} - \frac{x(3x - 4) \times 3(1 + 0) \times \sqrt{x - 1} }{2} }{2 {(x - 1)}^{3} } \\ \sf \: \: \: \: \: \: \: \: \: = \frac{ {(x - 1)}^{ \frac{3}{2} }(6x - 4) - 3 \sqrt{x - 1} \times x \times (3x - 4) }{2 {(x - 1)}^{3} } \\\sf \: \: \: \: \: \: \: \: \: = \: answer$