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En esta actividad obtendrás derivadas de funciones algebraicas, aplicando las reglas de derivación, para consolidar tu aprendizaje sobre las derivadas de funciones.

Ejercicio de escritura

Utilizando las definiciones de Newton y de Fermat, determina la derivada de la función lineal $f\left( x \right) = mx + b$ en $x = {x_0}$

$f\left( x \right) = mx + b$ $f'\left( x \right)=$
m
Solución 1. Utilizando la definición de Newton Solución 2. Utilizando la definición de Fermat

$\begin{array}{l} f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\ f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {m\left( {x + h} \right) + b} \right] - \left[ {mx + b} \right]}}{h}\\ f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {mx + mh + b} \right] - \left[ {mx + b} \right]}}{h}\\ f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{mx + mh + b - mx - b}}{h}\\ f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{mh}}{h}\\ f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} m\\ f'\left( x \right) = m \end{array}$

Por lo tanto, $f'\left( {{x_0}} \right) = m$

 $\begin{array}{l} f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f(a)}}{{x - a}}\\ f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left[ {mx + b} \right] - \left[ {m{x_0} + b} \right]}}{{x - {x_0}}}\\ f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{mx + b - m{x_0} - b}}{{x - {x_0}}}\\ f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{mx - m{x_0}}}{{x - {x_0}}}\\ f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{m\left( {x - {x_0}} \right)}}{{x - {x_0}}}\\ f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} m\\ f'\left( {{x_0}} \right) = m \end{array}$
Ejercicio de selección

Calcula la derivada de la función $f\left( x \right) = \frac{{4{x^5} - 6}}{{3{x^2} + 5}}$ utilizando la regla de derivación correspondiente.

$\begin{array}{l} f\left( x \right) = \frac{{4{x^5} - 6}}{{3{x^2} + 5}}\\ \frac{d}{{dx}}f\left( x \right) = \frac{{\left( {3{x^2} + 5} \right)\frac{d}{{dx}}\left[ {4{x^5} - 6} \right] - \left( {4{x^5} - 6} \right)\frac{d}{{dx}}\left[ {3{x^2} + 5} \right]}}{{{{\left( {3{x^2} + 5} \right)}^2}}}\\ \frac{d}{{dx}}f\left( x \right) = \frac{{\left( {3{x^2} + 5} \right)\left( {20{x^4}} \right) - \left( {4{x^5} - 6} \right)\left( {6x} \right)}}{{{{\left( {3{x^2} + 5} \right)}^2}}}\\ \frac{d}{{dx}}f\left( x \right) = \frac{{60{x^6} + 100{x^4} - 24{x^6} + 36x}}{{{{\left( {3{x^2} + 5} \right)}^2}}}\\ \frac{d}{{dx}}f\left( x \right) = \frac{{36{x^6} + 100{x^4} + 36x}}{{{{\left( {3{x^2} + 5} \right)}^2}}} \end{array}$

Calcula la derivada de la función $f\left( x \right) = \left( {\sqrt x + 2x} \right)\left( {{x^2} + 3x} \right)$ utilizando las reglas de derivación.

$\begin{array}{l} f\left( x \right) = \left( {\sqrt x + 2x} \right)\left( {{x^2} + 3x} \right)\\ f\left( x \right) = \left( {{x^{\frac{1}{2}}} + 2x} \right)\left( {{x^2} + 3x} \right)\\ \frac{d}{{dx}}f\left( x \right) = \left( {{x^{\frac{1}{2}}} + 2x} \right)\frac{d}{{dx}}\left( {{x^2} + 3x} \right) + \left( {{x^2} + 3x} \right)\frac{d}{{dx}}\left( {{x^{\frac{1}{2}}} + 2x} \right)\\ \frac{d}{{dx}}f\left( x \right) = \left( {{x^{\frac{1}{2}}} + 2x} \right)\left( {2x + 3} \right) + \left( {{x^2} + 3x} \right)\left( {\frac{1}{2}{x^{\frac{1}{2} - 1}} + 2} \right)\\ \frac{d}{{dx}}f\left( x \right) = \left( {{x^{\frac{1}{2}}} + 2x} \right)\left( {2x + 3} \right) + \left( {{x^2} + 3x} \right)\left( {\frac{1}{2}{x^{ - \frac{1}{2}}} + 2} \right)\\ \frac{d}{{dx}}f\left( x \right) = 2{x^{\frac{3}{2}}} + 3{x^{\frac{1}{2}}} + 4{x^2} + 6x + \frac{1}{2}{x^{\frac{3}{2}}} + 2{x^2} + \frac{3}{2}{x^{\frac{1}{2}}} + 6x\\ f\left( x \right) = 4{x^2} + 2{x^2} + 2{x^{\frac{3}{2}}} + \frac{1}{2}{x^{\frac{3}{2}}} + 6x + 6x + 3{x^{\frac{1}{2}}} + \frac{3}{2}{x^{\frac{1}{2}}}\\ f\left( x \right) = \frac{5}{2}{x^{\frac{3}{2}}} + \frac{9}{2}{x^{\frac{1}{2}}} + 6{x^2} + 12x\\ f\left( x \right) = \frac{5}{2}{x^{\frac{2}{2}}}{x^{\frac{1}{2}}} + \frac{9}{2}{x^{\frac{1}{2}}} + 6{x^2} + 12x\\ f\left( x \right) = \frac{5}{2}x\sqrt x + \frac{9}{2}\sqrt x + 6{x^2} + 12x \end{array}$

Calcula la derivada de la función $f\left( x \right) = \sqrt {\frac{{x - 2}}{{x + 3}}} $

$f\left( x \right) = \sqrt {\frac{{x - 2}}{{x + 3}}} $

$y = \sqrt {\frac{{x - 2}}{{x + 3}}}$ donde $u = \frac{{x - 2}}{{x + 3}}$; entonces $y = \sqrt u = {u^{\frac{1}{2}}}$

\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} = \frac{d}{{du}}y \cdot \frac{d}{{dx}}u\\ \frac{{dy}}{{dx}} = \frac{d}{{du}}\left[ {{u^{\frac{1}{2}}}} \right] \cdot \frac{d}{{dx}}\left[ {\frac{{x - 2}}{{x + 3}}} \right]\\ \frac{{dy}}{{dx}} = \frac{1}{2}{u^{\frac{1}{2} - 1}} \cdot \frac{{\left( {x + 3} \right)\frac{d}{{dx}}\left[ {x - 2} \right] - \left( {x - 2} \right)\frac{d}{{dx}}\left[ {x + 3} \right]}}{{{{\left( {x + 3} \right)}^2}}}\\ \frac{{dy}}{{dx}} = \frac{1}{2}{u^{ - \frac{1}{2}}} \cdot \frac{{\left( {x + 3} \right)\left( {x - 2} \right)\left( 1 \right)}}{{{{\left( {x + 3} \right)}^2}}}\\ \frac{{dy}}{{dx}} = \frac{1}{2}{u^{ - \frac{1}{2}}} \cdot \frac{{\left( {x + 3} \right)\left( {x - 2} \right)\left( 1 \right)}}{{{{\left( {x + 3} \right)}^2}}}\\ \frac{{dy}}{{dx}} = \frac{1}{2}{u^{ - \frac{1}{2}}} \cdot \frac{5}{{{{\left( {x + 3} \right)}^2}}} = \frac{1}{{2{u^{\frac{1}{2}}}}} \cdot \frac{5}{{{{\left( {x + 3} \right)}^2}}} = \frac{1}{{2\sqrt u }} \cdot \frac{5}{{{{\left( {x + 3} \right)}^2}}}\\ \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {\frac{{x - 2}}{{x + 3}}} }} \cdot \frac{5}{{{{\left( {x + 3} \right)}^2}}} = \frac{5}{{2 \cdot \frac{{\sqrt {x - 2} }}{{\sqrt {x + 3} }} \cdot {{\left( {x + 3} \right)}^2}}}\\ \frac{{dy}}{{dx}} = \frac{{5 \cdot \sqrt {x + 3} }}{{2 \cdot \sqrt {x - 2} \cdot {{\left( {x + 3} \right)}^2}}} \end{array}