Cal. Diferencial Sec. 2.5

Regla de la cadena

En los problemas del 1 al 61 derivar la función indicada. Las letras \(\boldsymbol{a}\), \(\boldsymbol{b}\) y \(\boldsymbol{c}\) denotan constantes.

\[ y = \left( x^2 - 3x + 5 \right)^3 \]
\[ f(x) = (15 - 8x)^4 \]
\[ g(t) = \left( 2t^3 - 1 \right)^{-3} \]
\[ z = \frac{ 1 }{ \left( 5x^5 - x^4 \right)^8 } \]
\[ y = (3x^2 - 8)^3 \left( -4x^2 + 1 \right)^4 \]
\[ f(u) = \frac{ 2u^3 + 1 }{ u^2 - 1 } \]
\[ y = \left( \frac{ x-1 }{ x+3 } \right)^2 \]
\[ g(t) = \left( \frac{ 3t^2 + 2 }{ 2t^3 - 1 } \right)^2 \]
\[ y = \sqrt{ 1 - 2x } \]
\[ u = \sqrt{ 1 + t - 2 t^2 - 8 t^3 } \]
\[ h(x) = x^2 \sqrt{ x^4 - 1 } \]
\[ g( x ) = \frac{ x }{ \sqrt{x^2 + 1} } \]
\[ y = \sqrt{3x^2 - 1} \sqrt[3]{ 2x + 1 } \]
\[ z = \left( 1 - 3x^2 \right)^2 \left( \sqrt{x} + 1 \right)^{-2} \]
\[ h(t) = \frac{ 1 + t }{ \sqrt{1 - t} } \]
\[ z = \sqrt[3]{ \frac{1}{ 1 + t^2 } } \]
\[ z = \sqrt[3]{ b + ax^3 } \]
\[ f(x) = \frac{ x }{ b^2 \sqrt{ b^2 + x^2 } } \]
\[ y = \frac{ 1 - \sqrt{1 + x} }{ 1 + \sqrt{1 + x} } \]
\[ f(x) = \sqrt{ (x - a)(x - b)(x - c) } \]
\[ y = \sqrt[3]{ x + \sqrt{x} } \]
\[ y = \sqrt{ x + \sqrt{ x + \sqrt{x} } } \]
\[ y = \tan 4x \]
\[ y = 2 \cot \frac{x}{2} \]
\[ u = \cos \left( x^3 \right) \]
\[ y = \cos^3 x \]
\[ y = \tan \left( x^4 \right) + \tan^4 x \]
\[ z = \cos \sqrt{x} \]
\[ u = \sqrt{ \cos x } \]
\[ y = \sqrt{ \cos \sqrt{x} } \]
\[ y = \sqrt[3]{ \tan 3x } \]
\[ y = \cot \sqrt[3]{ 1 + x^2 } \]
\[ y = \frac{4}{ \sqrt{ \sec x } } \]
\[ y = \text{cosec } \frac{1}{x^2} \]
\[ y = \text{ sen }^3 \left[ \frac{1 - \sqrt{x}}{ 1 + \sqrt{x} } \right] \]
\[ y = \frac{ \tan x }{ \sqrt{ \sec^2 (x) + 1 } } \]
\[ y = \sqrt{ \frac{ 1 + \text{ sen } x }{ 1 - \text{ sen } x } } \]
\[ y = \sqrt{ 1 + \cot \left( x + \frac{1}{x} \right) } \]
\[ y = \frac{ \cot (x/2) }{ \sqrt{ 1 - \cot^2 ( x / 2 ) } } \]
\[ y = \sqrt{ a \text{ sen }^2 x + b \cos^2 x } \]
\[ y = \cos ( \cos x ) \]
\[ y = \text{ sen } \left( \cos x^2 \right) \]
\[ y = \text{ sen }^2 ( \cos 4x ) \]
\[ y = \text{ sen } \left( \text{ sen } ( \text{ sen } x ) \right) \]
\[ y = \cos^2 ( \cos x ) + \text{ sen }^2 ( \text{ sen } x ) \]
\[ y = \text{ sen } \left( \tan \sqrt{ \text{ sen } x } \right) \]
\[ y = \tan \left( \text{ sen }^2 x \right) \]
\[ y = \mathrm{e}^{ -3x^2 + 1 } \]
\[ y = 2^{ \sqrt{x} } \]
\[ y = x^n a^{ -x^2 } \]
\[ y = 3^{ \cot ( 1/t ) } \]
\[ y = 2^{ 3^{ \text{ sen }^2 x } } \]
\[ y = \sqrt{ \log_5 x } \]
\[ y = \ln \left( \frac{ x }{ \mathrm{e}^x } \right) \]
\[ y = \frac{ \ln t }{ \mathrm{e}^{ 2t } } \]
\[ y = \ln \frac{ \mathrm{e}^{4x} - 1 }{ \mathrm{e}^{4x} + 1 } \]
\[ y = \mathrm{e}^{ x \ln x } \]
\[ y = \ln \left( \frac{ x+1 }{ \sqrt{ x - 2 } } \right) \]
\[ y = \ln \left( \frac{ x + 1 }{ x - 1 } \right)^{ 3/5 } \]
\[ y = \ln \left( x^3 \text{ sen } x \right) \]
\[ y = \ln \cos \frac{x-1}{x} \]
Si \(G(x) = (g(x))^{2/3}\), \(g(2) = 125\) y \(g'(2) = 150\), hallar \(G'(2)\).
Si \(F(t) = \left[ f(\text{ sen } t) \right]^2\), \(f(0) = -3\) y \(f'(0) = 5\), hallar \(F'(0)\).
Dadas \(f(u) = \frac{1}{4}u^3 - 3u + 5\) y \(g(x) = \frac{x-1}{x+1}\), hallar la derivada de \(f \circ g\) de dos maneras:
1. Encontrando \((f \circ g)(x)\) y derivando este resultado.
2. Aplicando la regla de la cadena.

En los ejercicios del 65 al 69, hallar \(\boldsymbol{h'(x)}\) si:

\[ \boldsymbol{ h(x) = (f \circ g )(x) = f(g(x)). } \]

\[ f(u) = u^3 - 2u^2 - 5,\; g(x) = 2x - 1 \]

\[ \begin{aligned} &f(u) = u^3 - 2u^2 - 5, \\[1em] &\hspace{3em}g(x) = 2x - 1 \end{aligned} \]
\[ f(v) = \sqrt{v},\, g(x) = 2x^3 - 4 \]
\[ f(t) = t^5,\, g(x) = 1 - 2 \sqrt{x} \]
\[ f(u) = \frac{b-u}{b+u},\, g(x) = cx \]
\[ f(v) = \frac{1}{v}, \, g(x) = a \sqrt{ a^2 - x^2 } \]

En los ejercicios del 70 al 73 hallar \(\boldsymbol{\frac{dy}{dx}}\).

\[ y = 3u^3 - 4u^4 - 1,\, u = x^2 - 1 \]
\[ y = v^5,\, v = 3a + 2bx \]
\[ y = t^4,\, t = \frac{ ax + b }{ c } \]
\[ y = \frac{1}{ \sqrt{v} }, \, v = 3x^2 - 1 \]

En los ejercicios del 74 al 81, hallar la recta tangente y la recta normal al gráfico de la función dada en el punto \(\boldsymbol{(a,\, f(a))}\), para el valor especificado de \(\boldsymbol{a}\).

\[ f(x) = (2x^2 - 1)^3, \; a = -1 \]
\[ f(x) = \frac{3}{ \left( 2 - x^2 \right)^2 }, \; a=0 \]
\[ f(x) = \frac{ x-2 }{ \sqrt{ 3x + 6 } }, \; a=1 \]
\[ f(x) = \sqrt[3]{ x-1 }, \; a=-7 \]
\[ f(x) = \frac{ (x-1)^2 }{ (3x - 2)^2 }, \; a=\frac{1}{2} \]
\[ f(x) = \cot^2 x,\; a= \frac{\pi}{4} \]
\[ f(x) = \mid 1 - x^3 \mid, \; a=2 \]
\[ f(x) = \mid \text{ sen } 5x \mid, \; a = \frac{ \pi }{3} \]
Hallar las rectas tangentes al gráfico de \(f(x) = (x-1)(x-2)(x-3)\) en los puntos donde el gráfico corta al eje X.
Hallar los puntos en la gráfica de \(g(x) = x^2(x - 4)^2\) en los cuales la recta tangente es paralela al eje X.
Hallar las rectas tangentes al gráfico de \(f(x) = \frac{x-4}{x-2}\) en los puntos donde este gráfico corta a los ejes. ¿Qué particularidad tienen estas rectas?.
Hallar las rectas tangentes al gráfico de \(g(x) = \frac{x+4}{x+3}\) que pasan por el origen.
Hallar las rectas tangentes al gráfico de \(f(x) = 3x^2 - \ln x\) en el punto (1, 3).
Hallar las rectas tangentes al gráfico de \(y = \ln \left( 1 + \mathrm{e}^x \right)\) en el punto \((0, \, \ln 2)\).
Sean \(f\) y \(g\) dos funciones diferenciables tales que \(f'(u) = \frac{1}{u}\) y \(f(g(x)) = x\). Probar que \(g'(x) = g(x)\).

Ver Respuestas

Respuestas

\[ \frac{dy }{dx} = 3 \left( x^2 – 3x + 5 \right)^2 (2x – 3) \]
\[ f'(x) = -32 (15 – 8x)^3 \]
\[ g'(t) = – 18t^2 \left( 2 t^3 – 1 \right)^{-4} \]
\[ \frac{dz}{dx} = -8 \left( 5x^5 – x^4 \right)^{-9} \left( 25 x^4 – 4 x^3 \right) \]

\[ \frac{dz}{dx} = -8 \left( 5x^5 – x^4 \right)^{-9} \left( 25 x^4 – 4 x^3 \right) \]
\[ \frac{dy}{dx} = -32x \left( 3x^2 – 8 \right)^3 \left( -4x^2 + 1 \right)^3 + 18x \left( 3x^2 – 8 \right)^2 \left( -4x^2 + 1 \right)^4 \]

\[ \begin{aligned} \frac{dy}{dx} = &-32x \left( 3x^2 – 8 \right)^3 \left( -4x^2 + 1 \right)^3 \\[.5em] & + 18x \left( 3x^2 – 8 \right)^2 \left( -4x^2 + 1 \right)^4 \end{aligned} \]
\[ f'(u) = \frac{ 2u \left( u^3 – 3u – 1 \right) }{ \left( u^2 – 1 \right)^2 } \]
\[ \frac{dy}{dx} = \frac{8(x – 1)}{(x + 3)^3} \]
\[ g'(t) = \frac{ -12t \left( 3 t^2 + 2 \right) \left( t^3 + 2t + 1 \right) }{ \left( 2 t^3 – 1 \right)^3 } \]

\[ g'(t) = \frac{ -12t \left( 3 t^2 + 2 \right) \left( t^3 + 2t + 1 \right) }{ \left( 2 t^3 – 1 \right)^3 } \]
\[ y’ = – \frac{1}{\sqrt{1 – 2x}} \]
\[ u’ = \frac{1 – 4t – 24 t^2}{ 2 \sqrt{ 1 + t – 2t^2 – 8t^3 } } \]
\[ h'(x) = \frac{ 2 x^5 }{ \sqrt{ x^4 – 1 } } + 2x \sqrt{ x^4 – 1 } \]
\[ g'(x) = \frac{ 1 }{ \left( x^2 + 1 \right)^{\frac{3}{2}} } \]
\[ y’ = \frac{ 2 \sqrt{ 3x^2 – 1 } }{ 3 \sqrt[3]{ (2x + 1)^2 } } + \frac{ 3 x \sqrt[3]{ 2x + 1 } }{ \sqrt{ 3 x^2 – 1 } } \]
\[ z’ = – \frac{ \left( 1 – 3x^2 \right)^2 }{ \sqrt{x} \left( \sqrt{x} + 1 \right)^3 } – \frac{ 12x \left( 1 – 3x^2 \right) }{ \left( \sqrt{x} + 1 \right)^2 } \]

\[ z’ = – \frac{ \left( 1 – 3x^2 \right)^2 }{ \sqrt{x} \left( \sqrt{x} + 1 \right)^3 } – \frac{ 12x \left( 1 – 3x^2 \right) }{ \left( \sqrt{x} + 1 \right)^2 } \]
\[ h'(t) = \frac{ 3 – t }{ 2 (1 – t)^{\frac{3}{2}} } \]
\[ z’ = – \frac{ 2t }{ 3 \left( 1 + t^2 \right)^{\frac{4}{3}} } \]
\[ z’ = \frac{ a x^2 }{ \sqrt[3]{ \left( b + ax^3 \right)^2 } } \]
\[ f'(x) = \frac{ 1 }{ \sqrt{ \left( b^2 + x^2 \right)^3 } } \]
\[ y’ = \frac{ -1 }{ \sqrt{ 1 + x } \left( 1 + \sqrt{1 + x} \right)^2 } \]
\[ f'(x) = \frac{ 3 x^2 – 2 (a + b + c)x + ab + ac + bc }{ 2 \sqrt{ (x – a)(x – b)(x – c) } } \]

\[ f'(x) = \frac{ 3 x^2 – 2 (a + b + c)x + ab + ac + bc }{ 2 \sqrt{ (x – a)(x – b)(x – c) } } \]
\[ y’ = \frac{ 1 + 2 \sqrt{x} }{ 6 \sqrt{x} \left( x + \sqrt{x} \right)^{\frac{2}{3}} } \]
\[ y’ = \frac{ 1 + 2 \sqrt{x} + 2 \sqrt{ x + \sqrt{x} } + 4 \sqrt{x} \sqrt{ x + \sqrt{x} } }{ 8 \sqrt{x} \sqrt{ x + \sqrt{x} } \sqrt{ x + \sqrt{ x + \sqrt{x} } } } \]

\[ y’ = \frac{ 1 + 2 \sqrt{x} + 2 \sqrt{ x + \sqrt{x} } + 4 \sqrt{x} \sqrt{ x + \sqrt{x} } }{ 8 \sqrt{x} \sqrt{ x + \sqrt{x} } \sqrt{ x + \sqrt{ x + \sqrt{x} } } } \]
\[ y’ = 4 \sec^2 4x \]
\[ y’ = – \text{cosec}^2 \frac{x}{2} \]
\[ u’ = – 3 x^2 \text{ sen } \left( x^3 \right) \]
\[ y’ = – 3 \text{ sen } x \cos^2 x \]
\[ y’ = 4 x^3 \sec^2 \left( x^4 \right) + 4 \tan^3 x \sec^2 x \]

\[ y’ = 4 x^3 \sec^2 \left( x^4 \right) + 4 \tan^3 x \sec^2 x \]
\[ z’ = -\frac{ 1 }{ 2 \sqrt{x} } \text{ sen } \sqrt{x} \]
\[ u’ = – \frac{ \text{ sen } x }{ 2 \sqrt{ \cos x } } \]
\[ y’ = – \frac{ \text{ sen } \sqrt{x} }{ 4 \sqrt{x} \sqrt{ \cos \sqrt{x} } } \]
\[ y’ = \frac{ \sec^2 3x }{ ( \tan 3x )^{\frac{2}{3}} } \]
\[ y’ = – \frac{ 2x \text{ cosec}^2 \left( \sqrt[3]{ 1 + x^2 } \right) }{ 3 \left( 1 + x^2 \right)^{\frac{2}{3}} } \]
\[ y’ = – \frac{ 2 \tan x }{ \sqrt{ \sec x } } \]
\[ y’ = \frac{ 2 }{ x^3 } \text{ cosec } \frac{1}{x^2} \cot \frac{1}{x^2} \]
\[ y’ = \frac{ -3 }{ \sqrt{x} \left( 1 + \sqrt{x} \right)^2 } \text{ sen}^2 \left[ \frac{ 1 – \sqrt{x} }{ 1 + \sqrt{x} } \right] \cos \left[ \frac{1 – \sqrt{x}}{ 1 + \sqrt{x} }\right] \]

\[ \begin{aligned} y’ =& \frac{ -3 }{ \sqrt{x} \left( 1 + \sqrt{x} \right)^2 } \\[.5em] &\hspace{2em} \times \text{ sen}^2 \left[ \frac{ 1 – \sqrt{x} }{ 1 + \sqrt{x} } \right] \\[.5em] &\hspace{4em} \times \cos \left[ \frac{1 – \sqrt{x}}{ 1 + \sqrt{x} }\right] \end{aligned} \]
\[ y’ = \frac{2 \sec^2 x}{ \left( \sec^2 x + 1 \right)^{\frac{3}{2}} } \]
\[ y’ = \frac{\cos x}{ (1 – \text{ sen } x)^2 } \sqrt{ \frac{1 – \text{ sen } x}{ 1 + \text{ sen } x } } \]
\[ y’ = \frac{ \left( 1 – x^2 \right) \text{ cosec}^2 \left( x + \frac{1}{x} \right) }{ 2 x^2 \sqrt{ 1 + \cot \left( x + \frac{1}{x} \right) } } \]
\[ y’ = – \frac{ \text{ cosec}^2 \frac{x}{2} }{ 2 \left( 1 – \cot^2 \frac{x}{2} \right)^{ \frac{3}{2} } } \]
\[ y’ = \frac{ (a – b) \text{ sen } 2x }{ 2 \sqrt{ a \text{ sen}^2 x + b \cos^2 x } } \]
\[ y’ = \text{ sen } x \text{ sen } (\cos x) \]
\[ y’ = -2x \text{ sen} x^2 \cos \left( \cos x^2 \right) \]
\[ y’ = – 4 \text{ sen } 4x \text{ sen } ( 2 \cos 4x ) \]
\[ y’ = \cos x \cos (\text{ sen } x) \cos (\text{ sen } ( \text{ sen } x )) \]

\[ y’ = \cos x \cos (\text{ sen } x) \cos (\text{ sen } ( \text{ sen } x )) \]
\[ y’ = \text{ sen } x \text{ sen } ( 2 \cos x ) + \cos x \text{ sen } ( 2 \text{ sen } x ) \]

\[ \begin{aligned} y’ = &\text{ sen } x \text{ sen } ( 2 \cos x ) \\[.5em] &\hspace{2em} + \cos x \text{ sen } ( 2 \text{ sen } x ) \end{aligned} \]
\[ y’ = \frac{ \cos x }{ 2 \sqrt{ \text{ sen } x } } \sec^2 \left( \sqrt{\text{ sen } x} \right) \cos \left( \tan \left( \sqrt{ \text{ sen } x } \right) \right) \]

\[ \begin{aligned} y’ =& \frac{ \cos x }{ 2 \sqrt{ \text{ sen } x } } \sec^2 \left( \sqrt{\text{ sen } x} \right) \\[.5em] &\hspace{2em} \times \cos \left( \tan \left( \sqrt{ \text{ sen } x } \right) \right) \end{aligned} \]
\[ y’ = \text{ sen } 2x \sec^2 \left( \text{ sen}^2 x \right) \]
\[ y’ = – 6 x \mathrm{e}^{- 3 x^2 + 1} \]
\[ y’ = \frac{ \ln 2 }{ 2 \sqrt{x} } 2^{ \sqrt{x} } \]
\[ y’ = x^{n – 1} a^{-x^2} \left( n – 2 x^2 \ln a \right) \]
\[ y’ = \frac{\ln 3}{t^2} \text{ cosec}^2 \left( \frac{1}{t} \right) 3^{ \cot \left( \frac{1}{t} \right) } \]
\[ y’ = ( \ln 2 ) ( \ln 3 ) \text{ sen } (2x) 3^{ \text{ sen}^2 x } 2^{ 3^{ \text{ sen}^2 x } } \]
\[ y’ = \frac{ 1 }{ 2 ( \ln 5 ) x \sqrt{ \log_5 x } } \]
\[ y’ = \frac{1}{x} – 1 \]
\[ y’ = \frac{1 – 2t \ln t}{ t \mathrm{e}^{2t} } \]
\[ y’ = \frac{ 8 \mathrm{e}^{4x} }{ \mathrm{e}^{8x} – 1 } \]
\[ y’ = \mathrm{e}^{x \ln x} (1 + \ln x) \]
\[ y’ = \frac{x – 5}{ 2 (x + 1) (x – 2) } \]
\[ y’ = – \frac{6}{ 5 \left( x^2 – 1 \right) } \]
\[ y’ = \frac{3}{x} + \cot x \]
\[ y’ = – \frac{1}{x^2} \tan \frac{x – 1}{x} \]
\[ G'(2) = 20 \]
\[ F'(0) = -30 \]
\[ (f \circ g)’ (x) = – \frac{ 3 \left( 3 x^2 + 10x + 3 \right)}{ 2 (x + 1)^4 } \]
\[ h'(x) = \left[ 3 u^2 – 4 u \right] (2) = 6 (2x – 1)^2 – 8 (2x – 1) = 24 x^2 – 40 x + 14 \]

\[ \begin{aligned} h'(x) &= \left[ 3 u^2 – 4 u \right] (2) \\[.5em] &= 6 (2x – 1)^2 – 8 (2x – 1) \\[.5em] &= 24 x^2 – 40 x + 14 \end{aligned} \]
\[ h'(x) = \frac{ 1 }{ 2 \sqrt{v} } \left( 6 x^2 \right) = \frac{3 x^2}{ \sqrt{ 2 x^3 – 4 } } \]
\[ h'(x) = 5 t^4 \left( – \frac{1}{ \sqrt{x} } \right) = \frac{ – 5 \left( 1 – 2 \sqrt{x} \right)^4 }{ \sqrt{x} } \]

\[ \begin{aligned} h'(x) &= 5 t^4 \left( – \frac{1}{ \sqrt{x} } \right) \\[.5em] &= \frac{ – 5 \left( 1 – 2 \sqrt{x} \right)^4 }{ \sqrt{x} } \end{aligned} \]
\[ h'(x) = \frac{ – 2 bc }{ (b + cx)^2 } \]
\[ h'(x) = \left( -\frac{1}{v^2} \right) \left( \frac{ – a x }{ \sqrt{ a^2 – x^2 } } \right) = \frac{ x }{ a \left( a^2 – x^2 \right)^{\frac{3}{2}} } \]

\[ \begin{aligned} h'(x) &= \left( -\frac{1}{v^2} \right) \left( \frac{ – a x }{ \sqrt{ a^2 – x^2 } } \right) \\[.5em] &= \frac{ x }{ a \left( a^2 – x^2 \right)^{\frac{3}{2}} } \end{aligned} \]
\[ \frac{dy}{dx} = \left( 9 u^2 – 16 u^3 \right) (2x) = 18x \left( x^2 – 1 \right)^2 – 32x \left( x^2 – 1 \right)^3 \]

\[ \begin{aligned} \frac{dy}{dx} &= \left( 9 u^2 – 16 u^3 \right) (2x) \\[.5em] &= 18x \left( x^2 – 1 \right)^2 – 32x \left( x^2 – 1 \right)^3 \end{aligned} \]
\[ \frac{dy}{dx} = 5 v^4 (2b) = 10 b ( 3a + 2bx )^4 \]
\[ \frac{dy}{dx} = 4 t^3 \left( \frac{a}{c} \right) = \frac{ 4a (ax + b)^3 }{ c^4 } \]
\[ \frac{dy}{dx} = \frac{-1}{ 2 v^{ \frac{3}{2}} } (6x) = \frac{ -3x }{ \left( 3 x^2 – 1 \right)^{ \frac{3}{2} } } \]
\[ 12x + y + 11 = 0, \, x – 12y + 13 = 0 \]

\[ \begin{aligned} &12x + y + 11 = 0, \\[.5em] &x – 12y + 13 = 0 \end{aligned} \]
\[ y = \frac{3}{4}, \; x = 0 \]
\[ 7x – 18y – 13 = 0, \, 54x + 21y – 47 = 0 \]

\[ \begin{aligned} &7x – 18y – 13 = 0, \\[.5em] &54x + 21y – 47 = 0 \end{aligned} \]
\[ x – 12y – 17 = 0, \, 12x + y + 86 = 0 \]
\[ 8x – y – 3 = 0, \, 2x + 16y – 17 = 0 \]
\[ y + 4x – 1 – \pi = 0, \, 16y – 4x + \pi – 16 = 0 \]

\[ \begin{aligned} &y + 4x – 1 – \pi = 0, \\[.5em] &16y – 4x + \pi – 16 = 0 \end{aligned} \]
\[ y – 12x + 17 = 0, \, 12y + x – 86 = 0 \]
\(6y + 15x – 5\pi + 3 \sqrt{3} = 0\), \(30y – 12x + 4\pi + 15 \sqrt{3} = 0\)
En \((1, \, 0)\): \(y – 2x + 2 = 0\);

En \((2, \, 0)\): \(y + x – 2 = 0\);

En \((3, \, 0)\): \(y – 2x + 6 = 0\).
\[ (0, \, 0), \, (4, \, 0), \, (2, \, 16) \]
\(2y – x – 4 = 0\), \(2y – x + 4 = 0\). Son paralelas
\[ y + x = 0, \, 9y + x = 0 \]
\[ y – 5x + 2 = 0 \]
\[ 2y – x – 2 \ln 2 = 0 \]