Question

(iii) Differentiate x.3X w.r.t. log(1 + 2x) step by step​

Answer 1

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Question : Differentiate wrt x log(1+2x)

$\pink{\huge{\underline{\underline{❤Answer❤}}}}$

$\: \\ \implies \displaystyle \bf{log(1 + 2x).....given \: equation}$

$\: \: \bullet \underline{\bf{differentiating \: both \: sides \: with \: respect \: to \: x}}$

$\: \implies \displaystyle \sf{ \frac{dy}{dx} = log(1 + 2x)}$

$\: \: \implies \displaystyle \sf{ \frac{dy}{dx} = \frac{1}{(1 + 2x)} \times 2}$

$\: \: \implies \displaystyle \sf{ \frac{dy}{dx} = \frac{2}{(1 + 2x)} }$

$\: \bigstar \boxed {\bf{answer \: will \: be \: \frac{1}{1 + 2x} }}$

$\: \bullet \underline{ \bf{additional \: information}}$

$\: \star \bf{ \frac{dy}{dx} logx = \frac{1}{x} }$

$\: \star \bf{ \frac{dy}{dx} log(1 + 2x) = \frac{1}{1 + 2x} \times \frac{dy}{dx} (2x})$

$\: \\ \ = \: \bf{\frac{2}{1 + 2x} }...proved \: by \: using \: chain \: rule$