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To find the derivative of the integral
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt, \][/tex]
we will use the Leibniz rule for differentiation under the integral sign with variable limits of integration.
The Leibniz rule states that if you have an integral of the form
[tex]\[ F(x) = \int_{a(x)}^{b(x)} f(t) \, dt, \][/tex]
then
[tex]\[ \frac{dF}{dx} = f(b(x)) \cdot \frac{db}{dx} - f(a(x)) \cdot \frac{da}{dx}. \][/tex]
Here, our integral is
[tex]\[ \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt. \][/tex]
Let's identify the functions involved:
- The integrand [tex]\( f(t) = \frac{\ln(t)}{t} \)[/tex].
- The upper limit of integration [tex]\( b(x) = x^3 \)[/tex].
- The lower limit of integration [tex]\( a(x) = x^2 \)[/tex].
Now we will differentiate the integral according to the Leibniz rule:
1. Evaluate the integrand [tex]\( f(t) \)[/tex] at the upper limit [tex]\( t = x^3 \)[/tex]:
[tex]\[ f(x^3) = \frac{\ln(x^3)}{x^3}. \][/tex]
2. Evaluate the integrand [tex]\( f(t) \)[/tex] at the lower limit [tex]\( t = x^2 \)[/tex]:
[tex]\[ f(x^2) = \frac{\ln(x^2)}{x^2}. \][/tex]
3. Differentiate the upper limit [tex]\( b(x) = x^3 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (x^3) = 3x^2. \][/tex]
4. Differentiate the lower limit [tex]\( a(x) = x^2 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (x^2) = 2x. \][/tex]
Putting everything together, according to the Leibniz rule:
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{\ln(x^3)}{x^3} \cdot 3x^2 - \frac{\ln(x^2)}{x^2} \cdot 2x. \][/tex]
Simplify the expression:
- Note that [tex]\(\ln(x^3) = 3 \ln(x)\)[/tex]:
[tex]\[ \frac{\ln(x^3)}{x^3} = \frac{3 \ln(x)}{x^3}. \][/tex]
- Similarly, [tex]\(\ln(x^2) = 2 \ln(x)\)[/tex]:
[tex]\[ \frac{\ln(x^2)}{x^2} = \frac{2 \ln(x)}{x^2}. \][/tex]
Substitute these back into the derivative expression:
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{3 \ln(x)}{x^3} \cdot 3x^2 - \frac{2 \ln(x)}{x^2} \cdot 2x. \][/tex]
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{3 \cdot 3 \ln(x) x^2}{x^3} - \frac{2 \cdot 2 \ln(x) x}{x^2}. \][/tex]
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{9 \ln(x) x^2}{x^3} - \frac{4 \ln(x) x}{x^2}. \][/tex]
[tex]\[ = 9 \frac{\ln(x)}{x} - 4 \frac{\ln(x)}{x}. \][/tex]
[tex]\[ = (9 - 4) \frac{\ln(x)}{x}. \][/tex]
[tex]\[ = 5 \frac{\ln(x)}{x}. \][/tex]
Therefore, the final expression is
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = 5 \frac{\ln(x)}{x}. \][/tex]
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt, \][/tex]
we will use the Leibniz rule for differentiation under the integral sign with variable limits of integration.
The Leibniz rule states that if you have an integral of the form
[tex]\[ F(x) = \int_{a(x)}^{b(x)} f(t) \, dt, \][/tex]
then
[tex]\[ \frac{dF}{dx} = f(b(x)) \cdot \frac{db}{dx} - f(a(x)) \cdot \frac{da}{dx}. \][/tex]
Here, our integral is
[tex]\[ \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt. \][/tex]
Let's identify the functions involved:
- The integrand [tex]\( f(t) = \frac{\ln(t)}{t} \)[/tex].
- The upper limit of integration [tex]\( b(x) = x^3 \)[/tex].
- The lower limit of integration [tex]\( a(x) = x^2 \)[/tex].
Now we will differentiate the integral according to the Leibniz rule:
1. Evaluate the integrand [tex]\( f(t) \)[/tex] at the upper limit [tex]\( t = x^3 \)[/tex]:
[tex]\[ f(x^3) = \frac{\ln(x^3)}{x^3}. \][/tex]
2. Evaluate the integrand [tex]\( f(t) \)[/tex] at the lower limit [tex]\( t = x^2 \)[/tex]:
[tex]\[ f(x^2) = \frac{\ln(x^2)}{x^2}. \][/tex]
3. Differentiate the upper limit [tex]\( b(x) = x^3 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (x^3) = 3x^2. \][/tex]
4. Differentiate the lower limit [tex]\( a(x) = x^2 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (x^2) = 2x. \][/tex]
Putting everything together, according to the Leibniz rule:
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{\ln(x^3)}{x^3} \cdot 3x^2 - \frac{\ln(x^2)}{x^2} \cdot 2x. \][/tex]
Simplify the expression:
- Note that [tex]\(\ln(x^3) = 3 \ln(x)\)[/tex]:
[tex]\[ \frac{\ln(x^3)}{x^3} = \frac{3 \ln(x)}{x^3}. \][/tex]
- Similarly, [tex]\(\ln(x^2) = 2 \ln(x)\)[/tex]:
[tex]\[ \frac{\ln(x^2)}{x^2} = \frac{2 \ln(x)}{x^2}. \][/tex]
Substitute these back into the derivative expression:
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{3 \ln(x)}{x^3} \cdot 3x^2 - \frac{2 \ln(x)}{x^2} \cdot 2x. \][/tex]
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{3 \cdot 3 \ln(x) x^2}{x^3} - \frac{2 \cdot 2 \ln(x) x}{x^2}. \][/tex]
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = \frac{9 \ln(x) x^2}{x^3} - \frac{4 \ln(x) x}{x^2}. \][/tex]
[tex]\[ = 9 \frac{\ln(x)}{x} - 4 \frac{\ln(x)}{x}. \][/tex]
[tex]\[ = (9 - 4) \frac{\ln(x)}{x}. \][/tex]
[tex]\[ = 5 \frac{\ln(x)}{x}. \][/tex]
Therefore, the final expression is
[tex]\[ \frac{d}{dx} \int_{x^2}^{x^3} \frac{\ln(t)}{t} \, dt = 5 \frac{\ln(x)}{x}. \][/tex]
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