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I got the equation $ \ln(x^2+2x+\frac{1}{2}) = x $ and I need to find the derivative of the function that created by comparing the equation to 0.

First try I did:

$\ln(x^2+2x+\frac{1}{2}) = x \Rightarrow \ln(x^2+2x+\frac{1}{2}) - x = 0 \Rightarrow f(x) = \ln(x^2+2x+\frac{1}{2}) - x \Rightarrow f'(x) = \frac{2x+2}{x^2+2x+\frac{1}{2}} -1 $

Second try I did:

$ \ln(x^2+2x+\frac{1}{2}) = x => e^x=x^2+2x+\frac{1}{2} => f(x)=x^2+2x+\frac{1}{2}-e^x => f'(x)=2x+2-e^x $

So my question is where is my mistake? Which one of the derivatives isn't correct? and why?

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    Both of your derivatives are correct. However, you are differentiating two different functions. So their derivatives are going to be different. Also, I don't understand why $\ln (x^2+2x+\frac{1}{2})=x$ creates a function, Isn't that a equation, not a function? – 2017-01-22
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    @S.C.b. Please clarify what are the two different functions? I started in both cases with the same equation.. later marked it as function.. – 2017-01-22

1 Answers 1

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From the first try we get $$f'(x) =\frac {2x+2-(x^2+2x+0.5)}{x^2+2x+0.5} =\frac {1.5-x^2}{x^2+2x+0.5} $$

From the second try we get, $$f'(x) =2x+2-(e^x) =(2x+2)-(x^2+2x+0.5) =1.5-x^2$$ which is clearly not equal to that obtained first.

The mistake in the second try is that we know that $$e^x =x^2+2x+0.5$$ $$\Rightarrow x^2+2x+0.5-e^x =0 \tag{1}$$ Now if we equate the LHS of $(1)$ as $f(x)$ then it boils down to $$f (x)=0 \implies f'(x)=0$$ which is clearly not what we want. So the first is correct. Hope it helps.