Question

Podrías ayudarme con las siguientes Derivadas:

6.
$y = x { }^{3}$
7.
$y = x {}^{3} - x {}^{2}$
8.
$y = \frac{4x {}^{3} - 16}{ x - 2}$
9.
$y = \frac{2x}{x - 1}$
10.
$y = (x - 1)(x {}^{2} + x + 1)$
11.
$f(x) = \frac{3}{x {}^{2} }$
12.
$f(x) = \frac{x {}^{2} - 1}{x {}^{2} + 1 }$
13.
$f(x) = \sqrt{x - 2}$

14.
$f(x) = \sqrt{x {}^{2} - 4 }$
​

Answer 1

Respuesta:

$6)\\y=x^{3}\\y'=3x^{3-1} \\y'=3x^{2} \\7)\\y=x^{3} -x^{2} \\y'=3x^{3-1} -2x^{2-1}\\y'=3x^{2} -2x^{}$

$8)\\y=\frac{4x^{3}-16}{x-2}\\ y=\frac{(4x^{3}-16)(x-2)-(4x^{3}-16)(x-2)}{(x-2)^{2} }\\ y'=\frac{(12x^{2}-0)(x-2)-(4x^{3}-16)(1-0)}{(x-2)^{2} }\\ y'=\frac{(12x^{2})(x-2)-(4x^{3}-16)(1)}{(x-2)^{2}}\\ y'=\frac{(12x^{3}-24x^{2} )-(4x^{3}-16)}{(x-2)^{2}}\\y'=\frac{12x^{3}-24x^{2}-4x^{3}+16}{(x-2)^{2}}\\y'=\frac{8x^{3}-24x^{2}+16}{(x-2)^{2}}$

9)

$y=\frac{2x}{x-1}\\y=\frac{(2x)*(x-1)-(2x)*(x-1)}{(x-1)^{2} }\\y'=\frac{(2)*(x-1)-(2x)*(1-0)}{(x-1)^{2} }\\y'=\frac{2x-2-2x}{(x-1)^{2} }\\y'=\frac{-2}{(x-1)^{2} }$

10)

$y=(x-1)(x^{2} +x+1)\\y'=(1-0)(x^{2} +x+1)+(x-1)(2x+1)\\y'=(1)(x^{2} +x+1)+(x-1)(2x+1)\\y'=(x^{2} +x+1)+(2x^{2}+x-2x-1)\\y'=(x^{2} +x+1)+(2x^{2}-x-1)\\y'=x^{2} +x+1+2x^{2}-x-1\\y'=3x^{2}$

11)

$f(x)=\frac{3}{x^{2} } \\f(x)=\frac{(3)*(x^{2})-(3)*(x^{2})}{(x^{2})^{2}}\\f'(x)=\frac{(0)*(x^{2})-(3)*(2x)}{(x^{2})^{2}}\\f'(x)=\frac{(0)-(6x)}{(x^{2})^{2}} \\f'(x)=\frac{-6x}{x^{4}} \\f'(x)=\frac{-6}{x^{3}}$

Aunque este ejercicio también se puede hacer usando las leyes de los exponentes, quedando así:

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

12)

$f(x)=\frac{x^{2}-1}{x^{2}+1} \\f(x)=\frac{(x^{2}-1)(x^{2}+1)-(x^{2}-1)(x^{2}+1)}{(x^{2}+1)^{2}}\\f'(x)=\frac{(2x-0)(x^{2}+1)-(x^{2} -1)(2x+0)}{(x^{2}+1)^{2}}\\f'(x)=\frac{(2x)(x^{2}+1)-(x^{2} -1)(2x)}{(x^{2}+1)^{2}}\\f'(x)=\frac{(2x^{3}+2x )-(2x^{3}-2x )}{(x^{2}+1)^{2}}\\f'(x)=\frac{2x^{3}+2x -2x^{3}+2x }{(x^{2}+1)^{2}} \\f'(x)=\frac{4x}{(x^{2}+1)^{2}}$

13)

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

14)

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