Question

me me alludan con el prosesimiento​

Answer 1

Respuesta:

En el Paso a Paso

Explicación paso a paso:

2.-

$y = 6.\sqrt{10-x^{5} }$

$y = 6.(10-x^{5})^{\frac{1}{2} }$

$y' = (6)'.(10-x^{5} )^{\frac{1}{2} } + 6. [(10-x^{5})^{\frac{1}{2} } ]'$

$y'= 3.(10-x^{5} )^{\frac{-1}{2} } .(-5x^{4} )$

$y' = \frac{-15x^{4} }{\sqrt{10 - x^{5} } }$

3.-

$y = \frac{-7}{x} - \frac{5}{x} + \frac{1}{x}$

$y = -7x^{-1} - 5x^{-1} + x^{-1}$

$y' = 7x^{-2} + 5x^{-2} - x^{-2}$

$y' = \frac{7}{x^{2} } + \frac{5}{x^{2} } - \frac{1}{x^{2} }$

4.-

$y = \frac{9}{(3x^{2} +8)^{2} }$

$y' = \frac{(9)'. [(3x^{2} +8)^{2}] - (9).[(3x^{2} +8)^{2}]' }{[(3x^{2} +8)^{2}]^{2} }$

$y'= \frac{- 18. (3x^{2} +8).(6x)}{(3x^{2} +8)^{4}}$

$y' = \frac{-108x}{(3x^{2} +8)^{3}}$

5.-

$y = x^{\frac{3}{4} } - \frac{5}{7}$

$y' = \frac{3}{4} . x^{\frac{3}{4} - 1 } - (\frac{5}{7})'$

$y' = \frac{3}{4} .x^{\frac{-1}{4} }$

6.-

$y = \frac{3}{\sqrt[3]{x} } - \frac{5}{2\sqrt{x} } \\\\y = 3.(x)^{\frac{-1}{3} } - \frac{5 x^({\frac{-1}{2}) } }{2}$

$y' = - x^{\frac{-4}{3} } + \frac{5x^{\frac{-3}{2} } }{4} \\\\y' = \frac{-1}{\sqrt[3]{x^{4} } } + \frac{5}{4\sqrt{x^3} }$

7.-

$y = -6x^{7} - 8x^{5} + x^{3}$

$y' = -42x^{6} - 40x^{4} + 3x^{2}$

8.-

$y = [(\frac{9}{2}x + 1)]^{\frac{-4}{3} } + 5x$

$y'= \frac{-4}{3} . (\frac{9}{2} x + 1)^{\frac{-7}{3} } .(\frac{9}{2} ) + 5\\\\y'= \frac{-6}{(\frac{9}{2} x +1)^{\frac{7}{3} } } + 5\\\\y'= \frac{-6}{\sqrt[3]{(\frac{9}{2} x + 1) ^{7} } } + 5$