How to solve derivative/limit of $f(x)=x\sqrt{4-{x}^{2}}$

Zion Wheeler
2022-06-22
Answered

How to solve derivative/limit of $f(x)=x\sqrt{4-{x}^{2}}$

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nuvolor8

Answered 2022-06-23
Author has **32** answers

I think your algebra could look more like:

$\begin{array}{rl}& \frac{(x+h)\sqrt{4-(x+h{)}^{2}}-x\sqrt{4-{x}^{2}}}{h}\cdot \frac{(x+h)\sqrt{4-(x+h{)}^{2}}+x\sqrt{4-{x}^{2}}}{(x+h)\sqrt{4-(x+h{)}^{2}}+x\sqrt{4-{x}^{2}}}\\ & =\frac{(x+h{)}^{2}(4-(x+h{)}^{2})-{x}^{2}(4-{x}^{2})}{h((x+h)\sqrt{4-(x+h{)}^{2}}+x\sqrt{4-{x}^{2}})}\end{array}$

This leaves no radicals in the numerator.

In the numerator, once this is multiplied out, all h-free terms will have canceled out through adding terms to their negatives.

Then you can factor h from the top and cancel the h in the denominator. Then it will be OK to just let $h\to 0$

$\begin{array}{rl}& \frac{(x+h)\sqrt{4-(x+h{)}^{2}}-x\sqrt{4-{x}^{2}}}{h}\cdot \frac{(x+h)\sqrt{4-(x+h{)}^{2}}+x\sqrt{4-{x}^{2}}}{(x+h)\sqrt{4-(x+h{)}^{2}}+x\sqrt{4-{x}^{2}}}\\ & =\frac{(x+h{)}^{2}(4-(x+h{)}^{2})-{x}^{2}(4-{x}^{2})}{h((x+h)\sqrt{4-(x+h{)}^{2}}+x\sqrt{4-{x}^{2}})}\end{array}$

This leaves no radicals in the numerator.

In the numerator, once this is multiplied out, all h-free terms will have canceled out through adding terms to their negatives.

Then you can factor h from the top and cancel the h in the denominator. Then it will be OK to just let $h\to 0$

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Let $f:\mathbb{R}\to \mathbb{R}$, $f(x)={e}^{{x}^{2}}$. Now, consider F an antiderivative of f.

Calculate: $\underset{x\to \mathrm{\infty}}{lim}\frac{xF(x)}{f(x)}$.

I couldn't find the antiderivative of f, and besides finding F, I have no other ideas.

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Calculate: $\underset{x\to \mathrm{\infty}}{lim}\frac{xF(x)}{f(x)}$.

I couldn't find the antiderivative of f, and besides finding F, I have no other ideas.

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I'm stumped on how to get the most general antiderivative, F(x) , of $f(x)={e}^{x}+3secx(tanx+secx)$.

First, I split the equation on addition, since $\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$

$F(x)=\int {e}^{x}dx+3\int secx(tanx+secx)dx$

Then I use the Substitution Rule. sec(x) seems like a good choice for u since it's the inner function of a composition and du will encompass the other x terms

$u=secx\Rightarrow du=(tanx+secx)dx$

Then I have $F(x)={e}^{x}+3\int (u)du\phantom{\rule{0ex}{0ex}}F(x)={e}^{x}+\frac{3{u}^{2}}{2}+C\phantom{\rule{0ex}{0ex}}F(x)={e}^{x}+\frac{3se{c}^{2}x}{2}+C$

However, the answer key says the correct answer is $F(x)={e}^{x}+3secx+3tanx+C$.

Can someone please point out where I've gone wrong here?

I'm stumped on how to get the most general antiderivative, F(x) , of $f(x)={e}^{x}+3secx(tanx+secx)$.

First, I split the equation on addition, since $\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$

$F(x)=\int {e}^{x}dx+3\int secx(tanx+secx)dx$

Then I use the Substitution Rule. sec(x) seems like a good choice for u since it's the inner function of a composition and du will encompass the other x terms

$u=secx\Rightarrow du=(tanx+secx)dx$

Then I have $F(x)={e}^{x}+3\int (u)du\phantom{\rule{0ex}{0ex}}F(x)={e}^{x}+\frac{3{u}^{2}}{2}+C\phantom{\rule{0ex}{0ex}}F(x)={e}^{x}+\frac{3se{c}^{2}x}{2}+C$

However, the answer key says the correct answer is $F(x)={e}^{x}+3secx+3tanx+C$.

Can someone please point out where I've gone wrong here?

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