veneciasp

Answered

2022-07-14

Need help with logarithmic differentitation

I have the expression

$y=\sqrt{{x}^{2}(x+1)(x+2)}.$

I have tried looking at videos but I still cannot arrive at the correct answer and don't know how to get there.

By the way, the correct answer is

${y}^{\prime}=\frac{4{x}^{2}+9x+4}{2\sqrt{(x+1)(x+2)}}.$

Please, help.

I have the expression

$y=\sqrt{{x}^{2}(x+1)(x+2)}.$

I have tried looking at videos but I still cannot arrive at the correct answer and don't know how to get there.

By the way, the correct answer is

${y}^{\prime}=\frac{4{x}^{2}+9x+4}{2\sqrt{(x+1)(x+2)}}.$

Please, help.

Answer & Explanation

Ordettyreomqu

Expert

2022-07-15Added 22 answers

$y=\sqrt{{x}^{2}(x+1)(x+2)}$

$\mathrm{ln}y=\mathrm{ln}x+\frac{1}{2}\mathrm{ln}(x+1)+\frac{1}{2}\mathrm{ln}(x+2)$

$\frac{{y}^{\prime}}{y}=\frac{1}{x}+\frac{1}{2}\frac{1}{x+1}+\frac{1}{2}\frac{1}{x+2}$

${y}^{\prime}=x\sqrt{(x+1)(x+2)}(\frac{1}{x}+\frac{1}{2}\frac{1}{x+1}+\frac{1}{2}\frac{1}{x+2})=x\sqrt{(x+1)(x+2)}\left(\frac{4{x}^{2}+9x+4}{2x(x+1)(x+2)}\right)$

which I believe works out to your answer.

$\mathrm{ln}y=\mathrm{ln}x+\frac{1}{2}\mathrm{ln}(x+1)+\frac{1}{2}\mathrm{ln}(x+2)$

$\frac{{y}^{\prime}}{y}=\frac{1}{x}+\frac{1}{2}\frac{1}{x+1}+\frac{1}{2}\frac{1}{x+2}$

${y}^{\prime}=x\sqrt{(x+1)(x+2)}(\frac{1}{x}+\frac{1}{2}\frac{1}{x+1}+\frac{1}{2}\frac{1}{x+2})=x\sqrt{(x+1)(x+2)}\left(\frac{4{x}^{2}+9x+4}{2x(x+1)(x+2)}\right)$

which I believe works out to your answer.

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