Question

differentiate
`\sqrt{ \frac{(x - 3)( {x}^(2) + 4) }{ {3x}^(2) + 4x + 5 } }`with respect to x ​

Answer 1

Explanation:

Let
`{ \green{ \tt{y = \sqrt{ \frac{(x - 3)( {x}^(2) + 4)}{3 {x}^(2) + 4x + 5}}}}}`

Take log on both sides

`{ \purple{ \sf{ log_(e)y = log_(e) [\frac{(x - 3)( {x}^(2) + 4}{3 {x}^(2) + 4x + 5}]^{ (1)/(2)}}} } { \to}{ \tt{ {eq}^(n) (1)}}`

This above expression is in the form of

`{ \boxed{ \red {\sf{ {x}^(n) = nx}}}}`

`{ \boxed{ \red{ \sf{ log( (m)/(n) ) = log \: m \: - log \: n}}}}`

Let's apply these two formulas to eqⁿ (1) then,

`{ \purple{ \sf{ log_(e)y = (1)/(2)[log(x - 3) + log( {x}^(2) + 4) - log( {3x}^(2) + 4x + 5) ]}}}`

differentiate with respect to x.

`{ \blue{ \sf{ (1)/(y) (dy)/(dx) = (1)/(2)[ (1)/(x - 3) (1 - 0) + \frac{1}{ {x}^(2) + 4 } (2x - 0) - \frac{1}{3 {x}^(2) - 4x + 5} (6x + 4 + 0)]}}}`

`{ \blue{ \sf{ (dy)/(dx) = y[ (1)/(2(x - 3)) + \frac{x}{ {x}^(2) + 4} - \frac{3x + 2}{ {3x}^(2) + 4x + 5 } ]}}}`

where
`{ \boxed{ \purple{ \sf{y = \sqrt{ \frac{(x - 3)( {x}^(2) + 4) }{3 {x}^(2) + 4x + 5}}}}} }`

Answer 2

`let \: \: y = \sqrt{ \frac{(x - 3)( {x}^(2) + 4) }{ {3x}^(2) + 4x + 5 } } \\ \\ y =( {{\frac{(x - 3)( {x}^(2) + 4) }{ {3x}^(2) + 4x + 5 } }})^{ (1)/(2) } \\ \\ taking \: \: logarathim \ \: both \: \: sides \: \: \\ \: we \: \:get ⇒ \: ln(y) = {(1)/(2) ln(\frac{(x - 3)( {x}^(2) + 4) }{ {3x}^(2) + 4x + 5 } }) \\ \\ ⇒2 ln(y) = ln(x - 3) + ln( {x}^(2) + 4) - ln(3 {x}^(2) ) + ln(4x) + ln(5) \\ differeciate \: \: both \: \: the \: \: sides \: \: we \: \: get \: \: \\ ⇒ 2 * (1)/(y) ((dy)/(dx) ) = (1)/(x - 3) + \frac{1}{ {x}^(2) + 4} .2x - \frac{1}{3 {x}^(2) } .6x + (1)/(4x) .4 + 0 \\ \\ ⇒ (2)/(y) * (dy)/(dx) = (1)/(x - 3) + \frac{2x}{ {x}^(2) + 4} - (2)/(x) + (1)/(x) \\ \\ ⇒ (dy)/(dx) = (y)/(2) . ((1)/(x - 3) + \frac{2x}{ {x }^(2) + 4} - (1)/(x) ) \\ \\ put \: \: the \: \: value \: \: of \: \: y \: \: we \: \: get \: \\ ⇒ (dy)/(dx) = \frac{\sqrt{ \frac{(x - 3)( {x}^(2) + 4) }{ {3x}^(2) + 4x + 5 } } \\ }{2} * ((1)/(x - 3) + \frac{2x}{ {x }^(2) + 4} - (1)/(x) )`

Explanation:

used formula

`{ ln(m) }^(n) = n \: ln \: m \\ \\ ln(mn) = ln(m) + ln(n) \\ \\ ln( (m)/(n) ) = ln(m) - ln(n) \\ \\ (d)/(dx) ( ln(x) ) = (1)/(x)`