(after Test 2)
Topics:
• Problem 7 from the WeBWorK assignment on Derivatives of inverse trig functions
My version of the problem: Find $\frac{\textrm{d}y}{\textrm{d}x}$ if
$y = \tan^{-1}(\sqrt{4x^{2} – 1})$
Recall that $\frac{\textrm{d}}{\textrm{d}x}\left(\tan^{-1}(x) = \frac{1}{1+x^{2}}$,
so
$\frac{\textrm{d}}{\textrm{d}u}\left(\tan^{-1}(u)\right) = \frac{1}{1+u^{2}}$
To find $\frac{\textrm{d}y}{\textrm{d}x}$ we need to use the Chain Rule:
$\frac{\textrm{d}y}{\textrm{d}x} = \frac{\textrm{d}}{\textrm{d}u}\left(\tan^{-1}(u)\right)\bigg|_{u = \sqrt{4x^{2} -1}}\cdot \frac{\textrm{d}}{\textrm{d}x}(\sqrt{4x^{2} – 1})$
We need to find the derivative of the inner function using the General Power Rule (special case of the Chain Rule):
$\frac{\textrm{d}}{\textrm{d}x}(\sqrt{4x^{2} – 1}) = \frac{\textrm{d}}{\textrm{d}x}\left[(4x^{2} – 1)^{\frac{1}{2}}\right]$
$= \frac{1}{2}(4x^{2} – 1)^{-\frac{1}{2}}\cdot 8x$
$= \frac{4x}{\sqrt{4x^{2} – 1}}$
Put this and the derivative of $\tan^{-1}(u)$ into the Chain Rule formula for $\frac{\textrm{d}y}{\textrm{d}x}$:
$\frac{\textrm{d}y}{\textrm{d}x} = \frac{\textrm{d}}{\textrm{d}u}\left(\tan^{-1}(u)\right)\bigg|_{u = \sqrt{4x^{2} -1}}\cdot \frac{\textrm{d}}{\textrm{d}x}(\sqrt{4x^{2} – 1})$
$= \frac{1}{1+u^{2}}\bigg|_{u = \sqrt{4x^{2} -1}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}}$
$= \frac{1}{1+(\sqrt{4x^{2} -1})^{2}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}}$
WeBWorK does not require that we simplify the answer any further than this (as long as we carefully type it in correctly! remember to use preview!) but this can be considerably simplified:
$ \frac{1}{1+(\sqrt{4x^{2} -1})^{2}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}} = \frac{1}{1+ 4x^{2} -1}\cdot \frac{4x}{\sqrt{4x^{2} – 1}} = \frac{1}{4x^{2}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}} = \frac{1}{x\sqrt{4x^{2} – 1}}$
So $\frac{\textrm{d}y}{\textrm{d}x} = \frac{1}{x\sqrt{4x^{2} – 1}}$
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Review of limits at infinity:
Remember that $\displaystyle \lim_{x\rightarrow \infty}f(x) = L$ means that as x grows without bound (goes off to the right), the height of the graph (value of f(x)) gets closer and closer to L,
and similarly $\displaystyle \lim_{x\rightarrow -\infty}f(x) = L$ means that as x goes in the negative direction without bound (goes off to the left), the height of the graph (value of f(x)) gets closer and closer to L.
If L is a number (not $\infty$ or $-\infty$), the fact that the limit exists means that there is a horizontal asymptote $y = L$ on that side.
$\displaystyle \lim_{x\rightarrow \infty}f(x) = \infty$ means that as x goes off to the right, the height of the graph (value of f(x)) grows without bound.
$\displaystyle \lim_{x\rightarrow \infty}f(x) = -\infty$ means that as x goes off to the right, the height of the graph (value of f(x)) goes in the negative direction without bound.
And similarly for limits as x approaches $-\infty$.
Here are some examples from basic functions you have already seen. Look at the graphs in Desmos which are linked here to see why these limits are what they are.
• $y = x^{2}$: graph
$ \lim_{x\rightarrow \infty}x^{2} = \infty$ and $\lim_{x\rightarrow -\infty}x^{2} = \infty$
• $y = x^{3}$: graph
$ \lim_{x\rightarrow \infty}x^{3} = \infty$ and $\lim_{x\rightarrow -\infty}x^{3} = -\infty$
• $y = e^{x}$: graph
$ \lim_{x\rightarrow \infty}e^{x} = \infty$ and $\lim_{x\rightarrow -\infty}e^{x} = 0$.
$y = e^{x}$ has a horizontal asymptote on the left, which is $y=0$ (The x-axis)
• $y = \ln(x)$: graph
$ \lim_{x\rightarrow \infty}\ln(x) = \infty$ (hard to see, because the height grows very slowly! But if you keep going to the right the height will eventually be greater than any number your chooose.) $\lim_{x\rightarrow -\infty}\ln(x)$ does not exist, because the domain of $\ln(x)$ is only the interval $(0, \infty)$.
• $y = \sin(x)$: graph
$ \lim_{x\rightarrow \infty}\sin(x)$ and $\lim_{x\rightarrow -\infty}\sin(x)$ do not exist, because $\sin(x)$ oscillates between 1 and -1 through all the number in between as you go off either to the right or to the left.
• $y = \frac{1}{x}$: graph
$ \lim_{x\rightarrow \infty}\frac{1}{x} = 0$ and $\lim_{x\rightarrow -\infty}\frac{1}{x} = 0$.
$y = \frac{1}{x}$ has a horizontal asymptote on both sides, which is $y=0$ (The x-axis)
