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

Sagot :

[tex]y=\frac { { { x }^{ 3 }\left( 2x+3 \right) }^{ \frac { 1 }{ 2 } } }{ { \left( x-2 \right) }^{ 2 } }[/tex]

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

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

[tex]\\ \\ \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) }[/tex]

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

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

[tex]\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }[/tex]

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

[tex]\\ \\ \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)[/tex]
Panoyin
y = ((x³)(2x + 3)^1/2)
              (x - 2)²
y = ((x³)(√(2x + 3))
       (x - 2)(x - 2)
y = ((√x^6)(√2x) + (√x^6)(√3))
             (x² - 2x - 2x + 4)
y = ((√2x^7) + (√3x^6))
           (x² - 4x + 4)
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