Question

if
$y = ({x}^{2} - 3x + 1) {}^{ \frac{5}{2} }$
find
$\frac{d {}^{2} y}{d {x}^{2} }$
at x = 0

​

Answer 1

Answer:

$\frac{15}{4} ( { {x}^{2} - 3x + 1)}^{ \frac{1}{2} } ({2x - 3})^{2} + 5( { {x}^{2} - 3x + 1})^{ \frac{3}{2} }$

Explanation:

$\frac{dy}{dx} = \frac{5}{2} { ( {x}^{2} - 3x + 1)} ^{ \frac{3}{2} } (2x - 3)$

$\frac{ {d}^{2}y }{d {x}^{2} } = \frac{15}{4} { ( {x}^{2} - 3x + 1)} ^{ \frac{1}{2} }(2x - 3)(2x - 3) + \frac{5}{2} {( {x }^{2} - 3x + 1) }^{ \frac{3}{2} } (2)$

$\frac{ {d}^{2}y }{d {x}^{2} } = \frac{15}{4}{ ( {x }^{2} - 3x + 1)}^{ \frac{1}{2} } ({2x - 3})^{2} + 5{( {x}^{2} - 3x + 1)}^{ \frac{3}{2} }$