Question

DERIVA LAS SIGUIENTES FUNCIONES APLICANDO LAS FORMULAS DE DERIVADAS​

Answer 1

Respuesta:

1.

$y'=(5)'\\y'=0$

2.

$y'=(\frac{3}{4} x)'\\y'=\frac{3}{4} (x)'\\y'=\frac{3}{4}$

3.

$f'(x)=(x^{5}')\\f'(x)=5(x^{5-1})\\f'(x)=5x^{4}$

4. (Se ve un poco borroso, espero anotarlo bien)

$f'(x)=(\frac{x^{5}}{7})'\\ f'(x)=\frac{1}{7}(x^{5})'\\f'(x)= \frac{1}{7}(5)(x^{4})\\f'(x)= \frac{5x^{4}}{7}$

5.

$f(x)=x^{\frac{1}{2}}\\f'(x)=\frac{1}{2}(x^{\frac{1}{2}-1} )\\f'(x)=\frac{1}{2}(x^{-\frac{1}{2}})\\f'(x)=\frac{1}{2\sqrt{x} }$

6.

$f'(x)=(x^{4}-5x^{3}+8x^{2}-x-6)'\\f'(x)=4x^{3}-15x^{2}+16x-1$

7.

$y=(5-3x^{2})^{\frac{1}{2}}\\y'=(\frac{1}{2})(5-3x^{2})^{\frac{1}{2}-1 }(-6x)\\y'=(-3x)(5-3x^{2})^{-\frac{1}{2}}\\y'=-\frac{3x}{\sqrt{5-3x^{2}} }$

8.

$y'=(3)(3x-4)^{2}(3) \\y'=9(3x-4)^{2}$

9.

$y=x^{3}(3x+1)\\y'=(x^{3})(3)+(3x+1)(3x^{2})\\y'=3x^{3}+9x^{3}+3x^{2}\\y'=12x^{3}+3x^{2}$

10.

$f(s)=\frac{s^{2}-2}{s^{2}-6s} \\f'(s)=\frac{(s^{2}-6s)(2s)-(s^{2}-2)(2s-6)}{(s^{2}-6s)^{2}} \\f'(s)=\frac{2s^{3}-12s^{2}-(2s^{3}-4s-6s^{2}+12)}{(s^{2}-6s)^{2}}\\f'(s)=\frac{2s^{3}-12s^{2}-2s^{3}+4s+6s^{2}-12}{(s^{2}-6s)^{2}}\\f'(s)=\frac{-6s^{2}+4s-12}{(s^{2}-6s)^{2}}$

Suerte!!