Let's apply the derivative to both sides with respect to x. We'll use the chain rule.
[tex]y = \ln\left(x + \sqrt{x^2+a^2}\right)\\\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(x + \sqrt{x^2+a^2}\right)\right]\\\\\\\frac{dy}{dx} = \frac{1}{x+\sqrt{x^2+a^2}}*\frac{d}{dx}\left[x + \sqrt{x^2+a^2}\right]\\\\\\\frac{dy}{dx} = \frac{1}{x+(x^2+a^2)^{1/2}}*\left(1 + 2x*\frac{1}{2}(x^2+a^2)^{-1/2}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+(x^2+a^2)^{1/2}}*\left(1 + x*\frac{1}{(x^2+a^2)^{1/2}}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+W}*\left(1 + x*\frac{1}{W}\right)\\\\\\[/tex]
where W = (x^2+a^2)^(1/2) = sqrt(x^2+a^2)
Let's simplify that a bit.
[tex]\frac{dy}{dx} = \frac{1}{x+W}*\left(1 + x*\frac{1}{W}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+W}*\left(\frac{W}{W} + \frac{x}{W}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+W}*\frac{W+x}{W}\\\\\\\frac{dy}{dx} = \frac{1}{W}\\\\\\\frac{dy}{dx} = \frac{1}{\sqrt{x^2+a^2}}\\\\\\[/tex]
This concludes the first part of what your teacher wants.
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Now onto the second part.
In this case, a = 3 so a^2 = 3^2 = 9.
Recall that if [tex]g(x) = \frac{d}{dx}\left[f(x)\right][/tex]
then [tex]\displaystyle \int g(x)dx = f(x)+C[/tex]
We can say that f(x) is the antiderivative of g(x). The C is some constant.
So,
[tex]\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+a^2}}dx = \ln\left(x + \sqrt{x^2+a^2}\right)+C\\\\\\\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+3^2}}dx = \ln\left(x + \sqrt{x^2+3^2}\right)+C\\\\\\\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+9}}dx = \ln\left(x + \sqrt{x^2+9}\right)+C\\\\\\[/tex]
Now let g(x) = ln(x + sqrt(x^2+9) ) + C
Then compute
- g(0) = ln(3)+C
- g(4) = ln(9) = ln(3^2) = 2*ln(3)+C
Therefore,
[tex]\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+9}}dx = \ln\left(x + \sqrt{x^2+9}\right)+C\\\\\\\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+9}}dx = g(x)\\\\\\\displaystyle \displaystyle \int_{0}^{4} \frac{1}{\sqrt{x^2+9}}dx = g(4) - g(0)\\\\\\\displaystyle \displaystyle \int_{0}^{4} \frac{1}{\sqrt{x^2+9}}dx = (2\ln(3)+C)-(\ln(3)+C)\\\\\\\displaystyle \displaystyle \int_{0}^{4} \frac{1}{\sqrt{x^2+9}}dx = \ln(3)\\\\\\[/tex]
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Extra info:
Interestingly, WolframAlpha says that the result is [tex]\sinh^{-1}\left(\frac{4}{3}\right)[/tex], but we can rewrite that into ln(3) because inverse hyperbolic sine is defined as
[tex]\sinh^{-1}(x) = \ln\left(x + \sqrt{x^2+1}\right)[/tex]
which is the function your teacher gave you, but now a = 1.
If you plugged x = 4/3 into the hyperbolic sine definition above, then you should get [tex]\sinh^{-1}\left(\frac{4}{3}\right) = \ln\left(3\right)[/tex]
