Question

derivata functiei f(x)=x la puterea 2 supra x la puterea 2 +4

Answer 1

$f(x) = \frac{ {x}^{2} }{ {x}^{2} + 4 }$

$f'(x) = ( \frac{ {x}^{2} }{ {x}^{2} + 4} )'$

$f'(x) = \frac{ ({x}^{2} )' \times ( {x}^{2} + 4) - {x}^{2} ( {x}^{2} + 4)'}{ {( {x}^{2} + 4)}^{2} }$

$f'(x) = \frac{2 {x}^{2 - 1} ( {x}^{2} + 4) - {x}^{2} [( {x}^{2} )' + 4']}{ {( {x}^{2} + 4)}^{2} }$

$f'(x) = \frac{2x( {x}^{2} + 4) - {x}^{2}(2 {x}^{2 - 1} + 0)}{ ( { {x}^{2} )}^{2} + 2 \times {x}^{2} \times 4 + {4}^{2} }$

$f'(x) = \frac{2x \times {x}^{2} + 2x \times 4 - {x}^{2} \times 2x}{ {x}^{2 \times 2} + 8 {x}^{2} + 16 }$

$f'(x) = \frac{2 {x}^{3} + 8x - 2 {x}^{3} }{ {x}^{4} + 8 {x}^{2} + 16 }$

$f'(x) = \frac{8x}{ {x}^{4} + 8 {x}^{2} + 16}$