Question Statement

Find the derivative of the function:

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

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Background and Explanation

This problem requires applying the chain rule for differentiation, along with the derivatives of logarithmic functions and square root functions. You will need to differentiate a composite function where the outer function is the natural logarithm and the inner function involves a sum of $x$ and a square root term.

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Solution

We are given: $y=\ln \left(x+\sqrt{x^2+1}\right)$

To find $\frac{dy}{dx}$, we apply the chain rule. Let $u=x+\sqrt{x^2+1}$, so $y=\ln (u)$.

First, recall that $\frac{d}{du}[\ln (u)]=\frac{1}{u}$.

Now, differentiate the inner function $u=x+\sqrt{x^2+1}$ with respect to $x$: $\frac{du}{dx}=\frac{d}{dx}(x)+\frac{d}{dx}\left(\sqrt{x^2+1}\right)$

For the square root term, use the chain rule again: $\frac{d}{dx}\left(\sqrt{x^2+1}\right)=\frac{d}{dx}\left((x^2+1{)}^{1/2}\right)=\frac{1}{2}(x^2+1{)}^{-1/2}\cdot \frac{d}{dx}(x^2+1)$

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

Therefore: $\frac{du}{dx}=1+\frac{x}{\sqrt{x^2+1}}$

Now apply the chain rule $\frac{dy}{dx}=\frac{1}{u}\cdot \frac{du}{dx}$:

$\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+1}}\cdot \left(1+\frac{x}{\sqrt{x^2+1}}\right)$

Simplify the expression in parentheses by finding a common denominator: $1+\frac{x}{\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}$

Substituting back: $\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+1}}\cdot \frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}$

The $(x+\sqrt{x^2+1})$ terms cancel: $\frac{dy}{dx}=\frac{1}{\sqrt{x^2+1}}$

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Derivative of Natural Logarithm: $\frac{d}{dx}[\ln (u)]=\frac{1}{u}\cdot \frac{du}{dx}$
• Derivative of Square Root: $\frac{d}{dx}[\sqrt{u}]=\frac{1}{2\sqrt{u}}\cdot \frac{du}{dx}$ or equivalently $\frac{d}{dx}[u^{1/2}]=\frac{1}{2}{u}^{-1/2}\cdot u^{\prime }$
• Algebraic Simplification: Combining fractions by finding common denominators to facilitate cancellation

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Summary of Steps

1. Identify the outer function ($\ln (u)$) and inner function ($u=x+\sqrt{x^2+1}$)
2. Differentiate the outer function with respect to $u$: $\frac{d}{du}[\ln (u)]=\frac{1}{u}$
3. Differentiate the inner function with respect to $x$:
   • $\frac{d}{dx}[x]=1$
   • $\frac{d}{dx}[\sqrt{x^2+1}]=\frac{x}{\sqrt{x^2+1}}$ using the chain rule
4. Apply the chain rule: $\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+1}}\cdot \left(1+\frac{x}{\sqrt{x^2+1}}\right)$
5. Simplify the second factor by combining fractions: $\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}$
6. Cancel the common factor $(x+\sqrt{x^2+1})$ to obtain the final result: $\frac{1}{\sqrt{x^2+1}}$