Question

$if \: y = \frac{ log_{e}(x) }{ {x}^{2} } then \: find \: the \: value \: of \: \frac{dy}{dx}$
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Answer 1

Here is ur answer:

Given that,

$y = \frac{ log_{e}(x) }{ {x}^{2} } \: \: then$

$\frac{dy}{dx} = \frac{d}{dx} ( \frac{ log_{e}(x) }{ {x}^{2} } ) = \frac{ {x}^{2} \frac{d}{dx}( log_{e}x) - log_{e}x. \frac{d( {x}^{2}) }{ {dx} {}^{} } }{( {x}^{2}) {}^{2} }$

$\frac{dy}{dx} = \frac{ {x}^{2} \binom{ \frac{1}{x} }{} - log_{e}x(2 {x}^{2 - 1} ) }{ {x}^{4} } = \frac{x - {2x log_{e}x } }{ {x}^{4} }$

$\frac{dy}{dx} = \frac{x}{ {x}^{4} } - \frac{2 xlog_{e}x }{ {x}^{4} } = \frac{1}{ {x}^{3} } - \frac{2 log_{e}x }{ {x}^{3} } = \frac{1}{ {x}^{3} } - \frac{ log_{e} {x}^{2} }{ {x}^{3} } = \frac{1 - log_{e} {x}^{2} }{ {x}^{3} }$

Hope this will help you ❤

Answer 2

dy/dx=1/x

When x=3, dy/dx=1/3 and y=ln3

.'. gradient of tangent at x=3 is 1/3

Use point gradient formula: y-y1=m(x-x1)

y-ln3=1/3.(x-3)

y=x/3 + ln3 - 1 or x-3y+ln3-3=0