Gonzalez Cox
12/26/2023 · Senior High School
Usando logaritmos, encuentren las derivadas de las siguientes funciones \( \begin{array}{lll}\text { a) } y=x^{x+1} & \text { b) } y=t^{\sqrt{t}} & \text { c) } y=(\sqrt{t})^{t} \\ \text { d) } y=(\ln x)^{x} & \text { e) } y=(\cos s)^{s} & \text { f) } y=(\tan x)^{x^{2}+1}\end{array} \)
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Para encontrar las derivadas de las funciones dadas usando logaritmos, aplicaremos la técnica de logaritmos implícitos. Esto implica tomar el logaritmo natural de ambos lados de la ecuación y luego derivar. Vamos a resolver cada caso uno por uno.
### a) \( y = x^{x+1} \)
1. Tomamos el logaritmo natural:
\[
\ln y = \ln(x^{x+1}) = (x+1) \ln x
\]
2. Derivamos ambos lados:
\[
\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}((x+1) \ln x)
\]
Usamos la regla del producto:
\[
\frac{d}{dx}((x+1) \ln x) = \ln x + (x+1) \cdot \frac{1}{x} = \ln x + 1 + \frac{1}{x}
\]
3. Entonces:
\[
\frac{1}{y} \frac{dy}{dx} = \ln x + 1 + \frac{1}{x}
\]
4. Multiplicamos por \( y \):
\[
\frac{dy}{dx} = y \left( \ln x + 1 + \frac{1}{x} \right) = x^{x+1} \left( \ln x + 1 + \frac{1}{x} \right)
\]
### b) \( y = t^{\sqrt{t}} \)
1. Tomamos el logaritmo natural:
\[
\ln y = \ln(t^{\sqrt{t}}) = \sqrt{t} \ln t
\]
2. Derivamos ambos lados:
\[
\frac{1}{y} \frac{dy}{dt} = \frac{d}{dt}(\sqrt{t} \ln t)
\]
Usamos la regla del producto:
\[
\frac{d}{dt}(\sqrt{t} \ln t) = \frac{1}{2\sqrt{t}} \ln t + \sqrt{t} \cdot \frac{1}{t} = \frac{1}{2\sqrt{t}} \ln t + \frac{1}{\sqrt{t}}
\]
3. Entonces:
\[
\frac{1}{y} \frac{dy}{dt} = \frac{1}{2\sqrt{t}} \ln t + \frac{1}{\sqrt{t}}
\]
4. Multiplicamos por \( y \):
\[
\frac{dy}{dt} = y \left( \frac{1}{2\sqrt{t}} \ln t + \frac{1}{\sqrt{t}} \right) = t^{\sqrt{t}} \left( \frac{1}{2\sqrt{t}} \ln t + \frac{1}{\sqrt{t}} \right)
\]
### c) \( y = (\sqrt{t})^{t} \)
1. Tomamos el logaritmo natural:
\[
\ln y = \ln((\sqrt{t})^{t}) = t \cdot \frac{1}{2} \ln t = \frac{t}{2} \ln t
\]
2. Derivamos ambos lados:
\[
\frac{1}{y} \frac{dy}{dt} = \frac{d}{dt}\left(\frac{t}{2} \ln t\right)
\]
Usamos la regla del producto:
\[
\frac{d}{dt}\left(\frac{t}{2} \ln t\right) = \frac{1}{2} \ln t + \frac{1}{2}
\]
3. Entonces:
\[
\frac{1}{y} \frac{dy}{dt} = \frac{1}{2} \ln t + \frac{1}{2}
\]
4. Multiplicamos por \( y \):
\[
\frac{dy}{dt} = y \left( \frac{1}{2} \ln t + \frac{1}{2} \right) = (\sqrt{t})^{t} \left( \frac{1}{2} \ln t + \frac{1}{2} \right)
\]
### d) \( y = (\ln x)^{x} \)
1. Tomamos el logarit
Quick Answer
a) \( y = x^{x+1} \): Derivative is \( y' = x^{x+1} \left( \ln x + 1 + \frac{1}{x} \right) \).
b) \( y = t^{\sqrt{t}} \): Derivative is \( y' = t^{\sqrt{t}} \left( \frac{1}{2\sqrt{t}} \ln t + \frac{1}{\sqrt{t}} \right) \).
c) \( y = (\sqrt{t})^{t} \): Derivative is \( y' = (\sqrt{t})^{t} \left( \frac{1}{2} \ln t + \frac{1}{2} \right) \).
d) \( y = (\ln x)^{x} \): Derivative is \( y' = (\ln x)^{x} \left( \frac{1}{x} \ln \ln x + \frac{1}{x} \right) \).
e) \( y = (\cos s)^{s} \): Derivative is \( y' = (\cos s)^{s} \left( -\sin s \ln \cos s + \frac{1}{\cos s} \right) \).
f) \( y = (\tan x)^{x^{2}+1} \): Derivative is \( y' = (\tan x)^{x^{2}+1} \left( \frac{2x \tan x + x^2}{\cos^2 x} \ln \tan x + \frac{x^2 + 1}{\cos^2 x} \right) \).
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