Question

cual es la derivada de y = sec (5x-7x^3)

Answer 1

aplicas fórmulas y derivas la parte de adentro del paréntesis quedando el siguiente resultado

Answer 2

Respuesta:

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$y = \sec(5x - 7 {x}^{3} )$

$y' = \frac{d}{dx} ( \sec(5x - 7 {x}^{3} ) )$

$y' = \frac{d}{dg} ( \sec(g) ) \times \frac{ {d} }{dx} (5x - 7 {x}^{3} )$

$y' = \tan(g) \sec(g) \times (5x - 7 \times 3 {x}^{2} )$

$y' = \tan(5x - 7 {x}^{3} ) \sec(5x - 7 {x}^{3} ) \times ( 5 - 7 \times 3 {x}^{2} )$

$y' = \frac{ \sin(5x - 7 {x}^{3} ) \times (5 - 21 {x}^{2} )}{ \cos(5x - 7 {x}^{3} ) }$

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