Question

The second order derivative of log(log x) will be​

Answer 1

Answer:

let y= log(logx)

y1=1/x log x

y2= -(1+logx)/(x+logx)^

Answer 2

➤Given :

⇒Y=log(logx)

➤To find :

⇒The second order derivative of log(logx)=?

➤Solution :

Consider a function y=f(x) where f(x) is a polynomial .

$\green \bigstar \boxed{ \red{ \tt{y = log(logx)}}}$

⇒Now, Differentiating both side with respect to x.

$\\ \implies \sf \: \: \frac{dy}{dx} = \frac{d(log(logx))}{dx} \\ \\ \ \\ \implies \sf \: \frac{dy}{dx} = \frac{1}{logx} \times \frac{d(logx)}{dx} \\ \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{1}{logx} \times \frac{1}{x} \\ \\ \\ \implies \boxed{ \red{ \sf{ \frac{dy}{dx} = \frac{1}{x.logx} }}}$

⇒Again Differentiating with respect to x :

$\\ \: \implies \sf \: \frac{dy}{dx} \bigg( \frac{dy}{dx} \bigg) = \frac{d}{dx} \bigg( \frac{1}{x.logx} \bigg)$

$\pink \bigstar \color{blue} \: \boxed{ \bf{ \red{by \: using \: quotient \: rule : }}}$

$\\ \\ \implies \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ \frac{d(1)}{dx}(x.logx) - \frac{d(x.logx)}{dx} \times 1 }{ {(x.logx)}^{2} } \\ \\ \\ \implies \sf\frac{ {d}^{2} y}{ {dx}^{2} } = \frac{0.(x.logx) - \frac{d(x.logx) \times 1}{dx} }{ {(x.logx)}^{2} } \\ \\ \\ \implies \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ \frac{ - d(x.logx)}{dx} }{ {(x.logx)}^{2} }$

$\pink \bigstar \color{blue} \boxed{ \bf \red{by \: using \: product \: rule : }}$

$\\ \implies \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ - \bigg \{ \frac{ d(x) }{dx} logx + \frac{d(logx)}{dx} .x \bigg \}}{ {(x.logx)}^{2} } \\ \\ \\ \implies \sf \: \frac{ {d}^{2} y}{ {dx}^{2} } = \frac{ \bigg \{1.logx + \frac{1}{x} \times x \bigg \}}{ {(x.logx)}^{2} } \\ \\ \\ \implies \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ - \bigg \{logx + 1 \bigg \}}{ {(x.logx)}^{2} } \\ \\ \\ \implies \sf \boxed {\bf{ \frac{ {d}^{2} y}{ {dx}^{2} } = \frac{ - \bigg \{ logx + 1 \bigg \}}{ {(x.logx)}^{2} } }}$

⇒Thus ,

$\red \bigstar\bf \: \: the \: second \: order \: derivative \: is \: \\ \boxed{ \red {\bf{ \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ - \bigg \{logx + 1 \bigg \}}{ {(x.logx)}^{2} } }}}$

Formulas :

$\pink\bigstar\green{\bf{Using \: Quotient \: Rule :}}$

As, $\boxed{\sf{\bigg(\dfrac{u}{v}\bigg)'=\dfrac{u'v-v'u}{v^{2}}}}$

where u =1 and v=xlogx

$\\ \pink\bigstar\green{\bf{Using \: product \: Rule : \: in \: xlogx}}$

$\boxed{\sf{(uv') =u'v+uv'}}$