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I've been trying to attempt this question but I keep getting the wrong answer.

For $y'$, I get

$$y' = \frac{3}{2}(1+\ln(4x))^{1/2}\times\frac{1}{x}$$

which I think can be rewritten as:

$$y' = \frac{3}{2x}(1 + \ln(4x))^{1/2}$$

I then sub the value of $x$ in and I get $3/0.5e^3$ and I think that's $6e^-3$.

But the correct answer is $12e^{-3}$.

Any help or pointing out any mistakes I've made will be super helpful.

Thank you.

Kenta S
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Tiffany
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  • @mfl If $y=\ln(4x)=\ln(x)+\ln(4)$, then $dy/dx=1/x$. – Kenta S Feb 09 '21 at 21:35
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    Better to revise your title and question using $$ math format. I realized you used it once in your question. Better to change the format of all mathematic parts using $$. And, when I read A^3/2, I assume $\frac{A^3}{2}$ since exponent has priority to division. However, it seems you meant $A^{3/2}$. – Nima S Feb 09 '21 at 21:37

1 Answers1

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See MathJax to format the question.

Did you evaluate $(1+\ln(4x))^{\frac{1}{2}}$ at $x=\frac{1}{4}{e^3}$?

Since $$\frac{dy}{dx}=\frac{3}{2x}(1+\ln(4x))^{\frac{1}{2}}$$

At $x=\frac{1}{4}{e^3}$ we have

$$\frac{3}{2(\frac{1}{4}{e^3})}(1+\ln(e^{3}))^{\frac{1}{2}}=\frac{3}{0.5\cdot e^{3}}\cdot2=12e^{-3}.$$

Alessio K
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