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Juli2301 [7.4K]
3 years ago
7

Use logarithmic differentiation to find dy/dx: y=((x^3)(2x+3)^1/2) / (x-2)^2

Mathematics
2 answers:
babymother [125]3 years ago
6 0
Y = <u>((x³)(2x + 3)^1/2)</u>
              (x - 2)²
y = (<u>(x³)(√(2x + 3))</u>
       (x - 2)(x - 2)
y = <u>((√x^6)(√2x) + (√x^6)(√3))
</u>             (x² - 2x - 2x + 4)
y = <u>((√2x^7) + (√3x^6))</u>
           (x² - 4x + 4)
sukhopar [10]3 years ago
4 0
y=\frac { { { x }^{ 3 }\left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } }{ { \left( x-2 \right)  }^{ 2 } }

\\ \\ { \left( x-2 \right)  }^{ 2 }\cdot y={ x }^{ 3 }{ \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  }

\\ \\ \ln { \left( { \left( x-2 \right)  }^{ 2 }\cdot y \right)  } =\ln { \left( { x }^{ 3 }{ \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ \ln { \left( { \left( x-2 \right)  }^{ 2 } \right)  } +\ln { y } =\ln { \left( { x }^{ 3 } \right)  } +\ln { \left( { \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ 2\ln { \left( x-2 \right)  } +\ln { y } =3\ln { x } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right)  }

\\ \\ \ln { y } =3\ln { x } -2\ln { \left( x-2 \right)  } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right)  }

\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }

\\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }  \right)

\\ \\ \frac { dy }{ dx } =\frac { { x }^{ 3 }\sqrt { 2x+3 }  }{ { \left( x-2 \right)  }^{ 2 } } \cdot \left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }  \right)
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