Difference Between Right and Wrong

WRONG: $\frac{1}{(x^{2}+1)\sin x}\cdot\left((x^{2}+1)\cos x+(\sin x)\cdot2x\right)=\cos x+2x=2x\cos x$

You have to factor or distribute before you can cancel terms.

CORRECT: $\frac{1}{(x^{2}+1)\sin x}\cdot\left((x^{2}+1)\cos x+(\sin x)\cdot2x\right)=$
$\frac{(x^{2}+1)\cos x}{(x^{2}+1)\sin x}+\frac{2x\cdot\sin x}{(x^{2}+1)\sin x}=\frac{\cos x}{\sin x}+\frac{2x}{x^{2}+1}=\cot x+\frac{2x}{x^{2}+1}$

WRONG: $\frac{1}{2xyy^{\prime}+y^{2}}=\frac{1}{2xyy^{\prime}}+\frac{1}{y^{2}}$

CORRECT: Well... avoid that mistake.

WRONG: $2\sin4x^{2}=8\sin x^{2}$

CORRECT: Well... avoid that mistake.

WRONG: $\frac{\sin(2x^{2})(4x)}{x}=\sin(2x^{2})(3x)$

CORRECT: $\frac{\sin(2x^{2})(4x)}{x}=\sin(2x^{2})(4)=4\sin(2x^{2})$

WRONG: $\cos0=0$

CORRECT: $\cos0=1$

WRONG: $\cos=1$

CORRECT: Er... cosine is a function, so it should be $\cos x$ or
$\cos u$ etc...

(Most) Final Exam Solutions

\#2b)

Most people did chain rule:

$f^{\prime}(x)=\frac{1}{(x^{2}+1)\sin x}\left(2x\cdot\sin x+(x^{2}+1)\cos x\right)$

Most errors resulted in not having parentheses, not knowing how to
do chain rule, or perhaps as simple as not knowing the derivative
of $\sin x$ is $\cos x$.

A simpler way to do the problem is by writing

$f(x)=\ln((x^{2}+1)\sin x)=\ln(x^{2}+1)+\ln(\sin x)$

So that

$f^{\prime}(x)=\frac{2x}{x^{2}+1}+\frac{1}{\sin x}\cos x=\frac{2x}{x^{2}+1}+\cot x$.

Note that doing it this way part of the problem comes for free provided
you did \#2a correctly.

A Requested Explanation

In general: $F(x)=\int_{a}^{f(x)}g(t)dt$. Then $F^{\prime}(x)=g(f(x))\cdot f^{\prime}(x)$.

Or perhaps more generally: $F(x)=\int_{f(x)}^{g(x)}h(t)dt$. Then
to compute the derivate we should first rewrite $F(x)$.

$F(x)=\int_{f(x)}^{a}h(t)dt+\int_{a}^{g(x)}h(t)dt=-\int_{a}^{f(x)}h(t)dt+\int_{a}^{g(x)}h(t)dt$

$F^{\prime}(x)=-h(f(x))\cdot f^{\prime}(x)+h(g(x))\cdot g^{\prime}(x)$.

Last Minute Thoughts

$e^{2x}+e^{x}\neq e^{3x}$
$e^{2x}\cdot e^{x}=e^{2x+x}=e^{3x}$
In general, $a^b\cdot a^c=a^{b+c}$

It is important you remember
$\int \frac{1}{1+u^2}du=\tan^{-1}(u)$

Know the derivative of all the trigonometric functions: sin(x), cos(x), tan(x), sec(x), csc(x), cot(x).

Some Requested Solutions and Help

1e) $\lim_{x\to0}e^{x\cos\left(e^{-1/x}\right)}$

It is enough to solve the following limit:

$\lim_{x\to0}x\cos\left(e^{-1/x}\right)$

Why is it enough to show this?

In any case, we observe that

$-|x|\leq x\cos\left(e^{-1/x}\right)\leq|x|$

Do you know why this is?

Please Ask Questions

Hi All,

With the final being cumulative, it's hard to give advice. The practice final seems to hit the major key points. Follow what I said in class regarding practice with the method of substitution and memorizing certain trigonometry identities and integrals/derivatives and even basic values. Less often but possible is determining if an integrand is an odd or even function to help in your computations. In any case, I hope you have learned good study habits by now and I wish you good luck, but do ask questions and I will help the best I can. This includes solution write-ups.

Your TA,
Tim