How do you differentiate $y = {log}_{x} 2$ $y=log_x2$ ?

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Answer 1 · 2017-07-11T13:29:11

$\frac{d y}{d x} = - \frac{ln 2}{x {(ln x)}^{2}}$ $(dy)/(dx)=-ln2/(x(lnx)^2)$

As $y = {log}_{x} 2$ $y=log_x2$ , we have $y = \frac{ln 2}{ln x}$ $y=ln2/lnx$

Hence assuming $g (f (x)) = \frac{1}{f (x)}$ $g(f(x))=1/(f(x))$ , where $f (x) = ln x$ $f(x)=lnx$

$\frac{d y}{d x} = ln 2 \times \frac{d g (x)}{d f (x)} \times \frac{d f}{d x}$ $(dy)/(dx)=ln2xx(dg(x))/(df(x))xx(df)/(dx)$

$= ln 2 \times (- \frac{1}{{(ln x)}^{2}}) \times \frac{1}{x}$ $=ln2xx(-1/(lnx)^2)xx1/x$

$= - \frac{ln 2}{x {(ln x)}^{2}}$ $=-ln2/(x(lnx)^2)$

How do you differentiate y=logx2y=log_x2 ?