y = "\\frac { x\u00b2\\sqrt {7x-14}} {(1+x)\u2074}"
ln(y) = ln [ "\\frac { x\u00b2\\sqrt {7x-14}} {(1+x)\u2074}" ]
Using properties of logarithm
ln(y) = ln(x²) + ln "\\sqrt {7x-14}" - ln(1+x)⁴
=> ln(y) = 2 ln x + "\\frac {1} {2}" ln (7x-14) - 4ln(1+x)
Differentiating with respect to x
"\\frac {1} {y}" "\\frac {dy} {dx}" = "\\frac {2} {x}" + "\\frac {7} {2(7x-14)}" - "\\frac {4} {1+x}"
So
"\\frac {dy} {dx}" = y [ "\\frac {2} {x}" + "\\frac {7} {2(7x-14)}" - "\\frac {4} {1+x}" ]
= "\\frac { x\u00b2\\sqrt {7x-14}} {(1+x)\u2074}" [ "\\frac {2} {x}" + "\\frac {7} {2(7x-14)}" - "\\frac {4} {1+x}" ]
Comments
In the solution ln means the natural logarithm, that is, the logarithm with the base e.
Does dat in means log in this question
Leave a comment