Ejercicios resueltos de regla del producto y cociente

Aplicando la regla del producto: $\frac{d}{dx}[uv] = u'v + uv'$

$u = x^3 \Rightarrow u' = 3x^2$

$v = e^x \Rightarrow v' = e^x$

$f'(x) = 3x^2 \cdot e^x + x^3 \cdot e^x = e^x(3x^2 + x^3) = x^2e^x(3 + x)$

$u = 2x + 1 \Rightarrow u' = 2$

$v = 3x^2 - 4 \Rightarrow v' = 6x$

$g'(x) = 2(3x^2 - 4) + (2x + 1)(6x) = 6x^2 - 8 + 12x^2 + 6x = 18x^2 + 6x - 8$

$u = \sin x \Rightarrow u' = \cos x$

$v = \cos x \Rightarrow v' = -\sin x$

$h'(x) = \cos x \cdot \cos x + \sin x \cdot (-\sin x) = \cos^2 x - \sin^2 x = \cos(2x)$

Regla del cociente: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$

$u = x^2 + 1 \Rightarrow u' = 2x$

$v = x - 2 \Rightarrow v' = 1$

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

$u = e^x \Rightarrow u' = e^x$

$v = \sin x \Rightarrow v' = \cos x$

$g'(x) = \frac{e^x \sin x - e^x \cos x}{\sin^2 x} = \frac{e^x(\sin x - \cos x)}{\sin^2 x}$

$u = x^3 \Rightarrow u' = 3x^2$

$v = x^2 + 4 \Rightarrow v' = 2x$

$h'(x) = \frac{3x^2(x^2+4) - x^3(2x)}{(x^2+4)^2} = \frac{3x^4+12x^2-2x^4}{(x^2+4)^2} = \frac{x^4+12x^2}{(x^2+4)^2} = \frac{x^2(x^2+12)}{(x^2+4)^2}$

Primero aplicamos la regla del cociente:

$u = x^2 \sin x$, $v = e^x$

Para $u'$ usamos la regla del producto:

$u' = 2x \sin x + x^2 \cos x$

$v' = e^x$

$f'(x) = \frac{(2x \sin x + x^2 \cos x)e^x - x^2 \sin x \cdot e^x}{(e^x)^2}$

$= \frac{e^x(2x \sin x + x^2 \cos x - x^2 \sin x)}{e^{2x}} = \frac{2x \sin x + x^2 \cos x - x^2 \sin x}{e^x}$

$= \frac{x(2\sin x - x\sin x + x\cos x)}{e^x} = \frac{x\sin x(2-x) + x^2\cos x}{e^x}$

Aplicamos la regla del producto:

$u = x + 1 \Rightarrow u' = 1$

$v = \frac{x^2}{x-1}$; su derivada, por la regla del cociente:

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

$g'(x) = 1 \cdot \frac{x^2}{x-1} + (x+1) \cdot \frac{x(x-2)}{(x-1)^2}$

$= \frac{x^2}{x-1} + \frac{x(x+1)(x-2)}{(x-1)^2} = \frac{x^2(x-1) + x(x+1)(x-2)}{(x-1)^2}$

Numerador: $u = (2x+1)(x^2-1)$

$u' = 2(x^2-1) + (2x+1)(2x) = 2x^2-2 + 4x^2+2x = 6x^2+2x-2$

Denominador: $v = 3x+2 \Rightarrow v' = 3$

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

Expandiendo el numerador: $(2x+1)(x^2-1) = 2x^3-2x+x^2-1 = 2x^3+x^2-2x-1$

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

Sean $f(x) = u(x) \cdot v(x)$. Queremos demostrar que $f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{u(x+h)v(x+h) - u(x)v(x)}{h}$

Sumamos y restamos $u(x+h)v(x)$:

$f'(x) = \lim_{h \to 0} \frac{u(x+h)v(x+h) - u(x+h)v(x) + u(x+h)v(x) - u(x)v(x)}{h}$

$= \lim_{h \to 0} \left[u(x+h)\frac{v(x+h)-v(x)}{h} + v(x)\frac{u(x+h)-u(x)}{h}\right]$

$= u(x)v'(x) + v(x)u'(x)$ (usando continuidad de $u$)

Sea $f(x) = \frac{u(x)}{v(x)}$. Queremos demostrar que $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

$f'(x) = \lim_{h \to 0} \frac{\frac{u(x+h)}{v(x+h)} - \frac{u(x)}{v(x)}}{h}$

$= \lim_{h \to 0} \frac{u(x+h)v(x) - u(x)v(x+h)}{hv(x+h)v(x)}$

Sumando y restando $u(x)v(x)$:

$= \lim_{h \to 0} \frac{u(x+h)v(x) - u(x)v(x) + u(x)v(x) - u(x)v(x+h)}{hv(x+h)v(x)}$

$= \lim_{h \to 0} \frac{v(x)\frac{u(x+h)-u(x)}{h} - u(x)\frac{v(x+h)-v(x)}{h}}{v(x+h)v(x)}$

$= \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$

$u = \sin x \Rightarrow u' = \cos x$

$v = \cos x \Rightarrow v' = -\sin x$

$\frac{d}{dx}[\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$ ✓