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I'm trying to differentiate the equation below but I fear there must have been an error made. I can't seem to reconcile to the correct answer. The problem comes from James Stewart's Calculus Early Transcendentals, 7th Ed., Page 223, Exercise 25.

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

My Answer:

Differentiate using the natural log rule: $$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(x+(1+x^2)^{1/2}\right)'$$

Now to differentiate the second term, note the chain rule applied and then simplification:

$$\left(x+(1+x^2)^{1/2}\right)'=1+\frac{1}{2}\cdot(1+x^2)^{-1/2}\cdot(2x)$$

$$1+\frac{1}{2}\cdot(1+x^2)^{-1/2}\cdot(2x)=1+\frac{x}{(1+x^2)^{1/2}}$$

Our expression is now:

$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(1+\frac{x}{(1+x^2)^{1/2}}\right)$$

Distribute the left term across the two right terms for my result:

$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)+\left(\frac{x}{\left(x+(1+x^2)^{1/2}\right)\left(1+x^2\right)^{1/2}}\right)$$ $$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)+\left(\frac{x}{\left(x(1+x^2)^{1/2}\right)+(1+x^2)^{1}}\right)$$

At this point I can see that if I simplify further by adding the fractions I'll still have too many terms, and it will get awfully messy. The answer per the book (below) has far fewer terms than mine. I'd just like to know where I've gone wrong in my algebra. Thank you for your help. Here's the correct answer:

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

Integreek
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baverso
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7 Answers7

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The function is $y=\ln(x+\sqrt{1+x^2})$, so $$ y'=\frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)= \frac{1}{x+\sqrt{1+x^2}}\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}= \frac{1}{\sqrt{1+x^2}} $$

I don't think your result is wrong per se: you just miss to notice some simplifications. Avoid distributing before factoring.

egreg
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    So you combined the $1+x/\sqrt{1+x^2}$ into one fraction, and then noticed that a factor canceled out. Since OP didn't do this, he didn't notice the factor that cancels. – Akiva Weinberger Dec 27 '16 at 00:03
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    @AkivaWeinberger Yes, the OP performed the multiplication too early. – egreg Dec 27 '16 at 00:06
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An alternative method is to use the relation $$\ln\left(x+\sqrt{x^2+1}\right)=\sinh^{-1}x=-i\sin^{-1}ix$$

So,

$$\begin{align}\frac{\mathrm d}{\mathrm dx}\ln\left(x+\sqrt{x^2+1}\right)&=\frac{\mathrm d}{\mathrm dx}(-i\sin^{-1}ix)\\&=-i\times\frac{i}{\sqrt{1-(ix)^2}}\\&=\frac1{\sqrt{1+x^2}}\end{align}$$

Integreek
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You wrote:

Distribute the left term across the two right terms

That is the same mistake that we see when we consider simplifying $\dfrac 3 4\times \dfrac 5 3.$ One can multiply: $$ \frac{3\times 5}{4\times 3} = \frac{15}{12} $$ but that is not a good way to simplify. Usually one should cancel before multiplying:$\require{cancel}$ $$ \frac{\bcancel 3} 4 \times \frac 5 {\bcancel 3}. $$ Do the same thing here: cancel before multiplying: \begin{align} \left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(1+\frac{x}{(1+x^2)^{1/2}}\right) & = \left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left( \frac{ (1+x^2)^{1/2} + x}{(1+x^2)^{1/2}} \right) \\[10pt] & =\left(\frac 1 {\cancel{x+(1+x^2)^{1/2}}}\right)\cdot\left( \frac{ \cancel{ (1+x^2)^{1/2} + x}}{(1+x^2)^{1/2}} \right) \end{align}

Michael Hardy
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Hint: Your function is the hyberbolic arcsin (or the inverse hyperbolic sin). In order to derivate such function one should use the theorem for inverse derivatives. That is $$(f^{-1}(x))'=\frac{1}{f^{'}(f^{-1}(x))}.$$ See the link for more information . https://en.m.wikipedia.org/wiki/Inverse_hyperbolic_function

Michael Hardy
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Beslikas Thanos
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Let $u=x+\sqrt{1+x^2}$. Then $$u'=1+\left[\frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot 2x\right]=1+\frac{x}{\sqrt{1+x^2}}=\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}.$$ Thus, $$y'=\frac{1}{u}\cdot u'=\frac{1}{x+\sqrt{1+x^2}}\cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}.$$

Juniven Acapulco
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You were almost there. You just missed the calculation. You got this:

$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(1+\frac{x}{(1+x^2)^{1/2}}\right)$$

You should just work with the right term:

$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(\frac{(1+x^2)^{1/2}+x}{(1+x^2)^{1/2}}\right)=\left(\frac{(1+x^2)^{1/2}+x}{[(1+x^2)^{1/2}+x](1+x^2)^{1/2}}\right)$$

Now just simplify the term $(1+x^2)^{1/2}+x$ and get:

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

Arnaldo
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Option:

$e^y=$

$x+(1+x^2)^{1/2};$

Differentiate both sides with respect to $x:$

$e^y(dy/dx)=$

$1+$

$(1/2)(1+x^2)^{-1/2}2x;$

Now divide by both sides by $e^y(\neq 0)$.

Used: Chain rule

Peter Szilas
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