Differentiate \(\displaystyle{y}=\sqrt{{\frac{{{1}+{2}{x}}}{{{1}-{2}{x}}}}}\) logarithmically \(\displaystyle{y}=\sqrt{{{\left\lbrace{\left\lbrace{1}+{2}{x}\right\rbrace}{o}{v}{e}{r}{\left\lbrace{1}-{2}{x}\right\rbrace}\right\rbrace}}}\) \(\displaystyle{\ln{{y}}}={\left\lbrace{1}{o}{v}{e}{r}{2}\right\rbrace}{\ln{{\left({1}+{2}{x}\right)}}}-{\left\lbrace{1}{o}{v}{e}{r}{2}\right\rbrace}{\ln{{\left({1}-{2}{x}\right)}}}\) \(\displaystyle{\left\lbrace\frac{{1}}{{y}}\right\rbrace}{\left\lbrace\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\right\rbrace}={\left\lbrace\frac{{1}}{{2}}\right\rbrace}\times{\left\lbrace\frac{{2}}{{{\left({1}+{2}{x}\right)}}}\right\rbrace}-{\left\lbrace\frac{{1}}{{2}}\right\rbrace}\times{\left\lbrace\frac{{-{2}}}{{{\left({1}-{2}{x}\right)}}}\right\rbrace}\)

Brielle James

Brielle James

Answered question

2022-04-14

Differentiate y=1+2x12x logarithmically
y=1+2x1-2x
lny=12ln(1+2x)-12ln(1-2x)
1/ydy/dx=1/2×2/(1+2x)-1/2×-2/(1-2x)
1/ydy/dx=1/(1+2x)+1/(1-2x)
dy/dx=(1-2x)+(1+2x)/(1+2x)(1-2x)×y
dy/dx=2(1+2x)1/2/(1+2x)(1-2x)(1-2x)1/2
dy/dx=2/(1+2x)1/2(1-2x)3/2
However the answer is:
dy/dx=2/(1-2x)(1-4x2)
Where have I went wrong?

Answer & Explanation

mislifola5vo

mislifola5vo

Beginner2022-04-15Added 11 answers

2(1+2x)12(12x)32=2(1+2x)12(12x)12(12x)=2(14x2)12(12x)

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