Question

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

Answer 1

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)

Answer 2

`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)`