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Sagot :
Sure, let's differentiate each function with respect to their independent variable.
### 8.1) Differentiating [tex]\( y = \ln |a t + 3| + \ln t \)[/tex] with respect to [tex]\( t \)[/tex]
1. Recall the properties and differentiation rules of logarithmic functions.
2. For [tex]\( \ln |a t + 3| \)[/tex], we use the chain rule:
[tex]\[ \frac{d}{dt}\left(\ln |a t + 3|\right) = \frac{d}{dt}\left(\ln u\right) \cdot \frac{du}{dt}, \quad \text{where} \ u = |a t + 3|. \][/tex]
Differentiating [tex]\( \ln u \)[/tex] gives:
[tex]\[ \frac{d}{dt}\left(\ln u\right) = \frac{1}{u}. \][/tex]
Now, we need [tex]\( \frac{du}{dt} \)[/tex] where [tex]\( u = |a t + 3| \)[/tex]. The absolute value function means we need to consider the derivative inside the absolute value:
[tex]\[ \frac{du}{dt} = a \cdot \frac{a t + 3}{|a t + 3|}. \][/tex]
Combining these:
[tex]\[ \frac{d}{dt}\left(\ln |a t + 3|\right) = \frac{a}{a t + 3}. \][/tex]
3. For [tex]\( \ln t \)[/tex], the derivative is straightforward:
[tex]\[ \frac{d}{dt}(\ln t) = \frac{1}{t}. \][/tex]
4. Combining these results:
[tex]\[ \frac{dy}{dt} = \frac{a}{a t + 3} + \frac{1}{t}. \][/tex]
### 8.2) Differentiating [tex]\( g(t) = 2^{\ln 2 t} + \ln e^{2 t} \)[/tex] with respect to [tex]\( t \)[/tex]
1. Consider the function [tex]\( 2^{\ln 2 t} \)[/tex]. This can be simplified using the properties of exponents and logarithms:
[tex]\[ 2^{\ln 2 t} = (2^{\ln 2})^t = e^{(\ln 2) \ln 2 t} = e^{(\ln 2)^2 t}. \][/tex]
Then, we differentiate:
[tex]\[ \frac{d}{dt} \left( e^{(\ln 2)^2 t} \right) = e^{(\ln 2)^2 t} \cdot (\ln 2)^2. \][/tex]
2. For [tex]\( \ln e^{2 t} \)[/tex], recall that [tex]\( \ln e^u = u \)[/tex]:
[tex]\[ \ln e^{2 t} = 2 t, \][/tex]
and its derivative:
[tex]\[ \frac{d}{dt} (2 t) = 2. \][/tex]
3. Combining these:
[tex]\[ \frac{dg}{dt} = e^{(\ln 2)^2 t} \cdot (\ln 2)^2 + 2. \][/tex]
To get the exact form of [tex]\( g(t) \)[/tex]:
[tex]\[ e^{(\ln 2)^2 t} \cdot (\ln 2)^2 \text{ can be written as } 2^{t \ln 2} \cdot (\ln 2)^2. \][/tex]
Combining all results:
[tex]\[ \frac{dy}{dt} = \frac{a}{a t + 3} + \frac{1}{t}, \quad \frac{dg}{dt} = 2^{t \ln 2} \cdot (\ln 2)^2 + 2. \][/tex]
### 8.1) Differentiating [tex]\( y = \ln |a t + 3| + \ln t \)[/tex] with respect to [tex]\( t \)[/tex]
1. Recall the properties and differentiation rules of logarithmic functions.
2. For [tex]\( \ln |a t + 3| \)[/tex], we use the chain rule:
[tex]\[ \frac{d}{dt}\left(\ln |a t + 3|\right) = \frac{d}{dt}\left(\ln u\right) \cdot \frac{du}{dt}, \quad \text{where} \ u = |a t + 3|. \][/tex]
Differentiating [tex]\( \ln u \)[/tex] gives:
[tex]\[ \frac{d}{dt}\left(\ln u\right) = \frac{1}{u}. \][/tex]
Now, we need [tex]\( \frac{du}{dt} \)[/tex] where [tex]\( u = |a t + 3| \)[/tex]. The absolute value function means we need to consider the derivative inside the absolute value:
[tex]\[ \frac{du}{dt} = a \cdot \frac{a t + 3}{|a t + 3|}. \][/tex]
Combining these:
[tex]\[ \frac{d}{dt}\left(\ln |a t + 3|\right) = \frac{a}{a t + 3}. \][/tex]
3. For [tex]\( \ln t \)[/tex], the derivative is straightforward:
[tex]\[ \frac{d}{dt}(\ln t) = \frac{1}{t}. \][/tex]
4. Combining these results:
[tex]\[ \frac{dy}{dt} = \frac{a}{a t + 3} + \frac{1}{t}. \][/tex]
### 8.2) Differentiating [tex]\( g(t) = 2^{\ln 2 t} + \ln e^{2 t} \)[/tex] with respect to [tex]\( t \)[/tex]
1. Consider the function [tex]\( 2^{\ln 2 t} \)[/tex]. This can be simplified using the properties of exponents and logarithms:
[tex]\[ 2^{\ln 2 t} = (2^{\ln 2})^t = e^{(\ln 2) \ln 2 t} = e^{(\ln 2)^2 t}. \][/tex]
Then, we differentiate:
[tex]\[ \frac{d}{dt} \left( e^{(\ln 2)^2 t} \right) = e^{(\ln 2)^2 t} \cdot (\ln 2)^2. \][/tex]
2. For [tex]\( \ln e^{2 t} \)[/tex], recall that [tex]\( \ln e^u = u \)[/tex]:
[tex]\[ \ln e^{2 t} = 2 t, \][/tex]
and its derivative:
[tex]\[ \frac{d}{dt} (2 t) = 2. \][/tex]
3. Combining these:
[tex]\[ \frac{dg}{dt} = e^{(\ln 2)^2 t} \cdot (\ln 2)^2 + 2. \][/tex]
To get the exact form of [tex]\( g(t) \)[/tex]:
[tex]\[ e^{(\ln 2)^2 t} \cdot (\ln 2)^2 \text{ can be written as } 2^{t \ln 2} \cdot (\ln 2)^2. \][/tex]
Combining all results:
[tex]\[ \frac{dy}{dt} = \frac{a}{a t + 3} + \frac{1}{t}, \quad \frac{dg}{dt} = 2^{t \ln 2} \cdot (\ln 2)^2 + 2. \][/tex]
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