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1. (25 points) Let \(f_a(x) = \begin{cases} x^a & \text{if } x > 0, \\ 0 & \text{if } x \leq 0. \end{cases}\)

For which values of \(a\) is \(f\) continuous at zero?

If \(f\) is continuous at zero, let's consider \(\lim_{x\to0^{+}}x^{a}=f(0)=0\)

Then \(\forall\varepsilon,\exists\delta,\left|x\right|<\delta:\left|x^{a}\right|<\varepsilon\),

Then we consider three cases
1. \(a>0\)
  
  Then \(|x|^{a}<\delta^{a}<\varepsilon\), we just take \(\delta<\sqrt[a]{\varepsilon}\) 2. \(a<0\)
  
  Then \(|x|^{a}>\delta^{a}\), but we need to take \(\delta\) to let \(|x^a|<\varepsilon\) which is impossible 3. \(a=0\)
  
  Impossible since \(x^0=1\)
Thus \(a>0\), \(f\) continuous at zero
For which values of \(a\) is \(f\) differentiable at zero? In this case, is the derivative function continuous?

We have \(f_a(x) = \begin{cases} x^a & \text{if } x > 0, \\ 0 & \text{if } x \leq 0. \end{cases}\) Then \(\lim_{x\to0}\frac{f\left(x\right)}{x}=\begin{cases}\lim_{x\to0}x^{a-1},x>0\\ 0,x\leq0\end{cases}\)

When \(a-1=0\), \(\lim_{x\to0}\frac{f\left(x\right)}{x}=\begin{cases}\lim_{x\to0}x^0=1,x>0\\ 0,x\leq0\end{cases}\)

Thus when \(a=1\), \(f\) is not differentiable

When \(a-1<0\), then \(\lim_{x\to0}\frac{f\left(x\right)}{x}=\begin{cases}\lim_{x\to0}x^{a-1}=\lim_{x\to0}\frac{1}{x^{1-a}}\neq0,x>0\\ 0,x<0\end{cases}\), thus also impossible since \(1-a>0\)

Thus when \(a<1\), \(f\) is not differentiable

When \(a-1>0\), \(\lim_{x\to0}\frac{f\left(x\right)}{x}=\begin{cases}\lim_{x\to0}x^{a-1}=0,x>0\\ 0,x<0\end{cases}\) which is possible

Thus when \(a>1\), \(f\) is differentiable

When \(a>1\), the differentiable function is \(f^{\prime}_{a}(x)=\begin{cases}ax^{a-1} & \text{if }x>0,\\ 0 & \text{if }x\leq0.\end{cases}\)

Since two function are continuous, then we only need to check whether \(x=0\) is continuous.

Let's consider \(\lim_{x\to0^{+}}x^{a}=f(0)=0\). Then \(\forall\varepsilon,\exists\delta,\left|x\right|<\delta:\left|ax^{a-1}\right|<\varepsilon\), then \(|ax^{a-1}|<a\delta^{a-1}<\varepsilon\), we only need to take \(\delta<\sqrt[a-1]{\frac{\varepsilon}{a}}\)

Thus the derivative function is continuous
For which values of \(a\) is \(f\) twice-differentiable?

Since \(f^{\prime}_{a}(x)=\begin{cases}ax^{a-1} & \text{if }x>0,\\ 0 & \text{if }x\leq0.\end{cases}\) and two function are differentiable, then we only need to check whether \(x=0\) is differentiable.

\(\lim_{x\to0}\frac{f^{\prime}\left(x\right)}{x}=\begin{cases}\lim_{x\to0}ax^{a-2},x>0\\ 0,x<0\end{cases}\), then if it is differentiable, then \(\lim_{x\to0}ax^{a-2}=0\Rightarrow\lim_{x\to0}x^{a-2}=0\Rightarrow a>2\)

Thus if \(a>2\), \(f\) is twice-differentiable.

2. (25 points) Compute the derivative of the following functions:

\[ f(x) = (\sin x)^{\cos x} \]

Solution:

\(f(x)=e^{\cos x\ln\sin x}\), then \(f^{\prime}(x)=e^{\cos x\ln\sin x}\cdot\left(\cos x\ln\sin x\right)^{\prime}\)

Then \(\left(\cos x\ln\sin x\right)^{\prime}=-\sin x\cdot\ln\sin x+\cos x\frac{1}{\sin x}\cdot\cos x=-\sin x\cdot\ln\sin x+\cot x\cdot\cos x\)

Thus \(f^{\prime}\left(x\right)=\left(\sin x\right)^{\cos x}\cdot\left(-\sin x\cdot\ln\sin x+\cot x\cdot\cos x\right)=\frac{\sin x^{\left(\cos x\right)-1}}{\cos^2x}-\sin x^{\left(\cos x\right)+1}\ln\left(\sin x\right)\)
\[ f(x) = x^{(2x+1)^x} \]

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Solution:

\(f(x)=x^{(2x+1)^{x}}=e^{\left(2x+1\right)^{x}\ln x}=e^{e^{x\ln\left(2x+1\right)}\ln x}\), then \(f^{\prime}(x)=e^{e^{x\left(\ln2x+1\right)}\ln x}\cdot\left({e^{x\ln\left(2x+1\right)}\ln x}\right)^{\prime}\)

Thus \(\left({e^{x\ln\left(2x+1\right)}\ln x}\right)^{\prime}=e^{x\ln\left(2x+1\right)}\cdot\left({x\ln\left(2x+1\right)}\right)^{\prime}\cdot\ln x+e^{x\ln\left(2x+1\right)}\cdot\frac{1}{x}\)

Then \(\left({x\ln\left(2x+1\right)}\right)^{\prime}=\ln\left(2x+1\right)+x\frac{1}{2x+1}\cdot2=\ln\left(2x+1\right)+\frac{2x}{2x+1}\)

Finally, \(f^{\prime}(x)=x^{(2x+1)^{x}}\cdot\left(e^{x\ln\left(2x+1\right)}\cdot\left(\ln\left(2x+1\right)+\frac{2x}{2x+1}\right)\cdot\ln x+e^{x\ln\left(2x+1\right)}\cdot\frac{1}{x}\right)=x^{(2x+1)^{x}}\cdot{\left(2x+1\right)^{x}}\cdot\left(\left(\ln\left(2x+1\right)+\frac{2x}{2x+1}\right)\cdot\ln x+\frac{1}{x}\right)\)
\[ f(x) = \sin(\sin(\sin(\sin(x \cos(x))))) \]

Solution:

\(f'(x)=\cos(\sin(\sin(\sin(x \cos(x)))))\cdot (\sin(\sin(\sin(x \cos(x)))))'\)

\((\sin(\sin(\sin(x \cos(x)))))'=\cos(\sin(\sin(x \cos(x))))\cdot (\sin(\sin(x \cos(x))))'\)

\((\sin(\sin(x \cos(x))))'=\cos(\sin(x \cos(x)))\cdot (\sin(x \cos(x)))'\)

\((\sin(x \cos(x)))'=\cos(x\cos x)\cdot (x\cos x)'\)

\((x\cos x)'=\cos x-x\sin x\)

Finally, \(f^{\prime}(x)=\cos(\sin(\sin(\sin(x\cos(x)))\cdot\cos(\sin(\sin(x \cos(x))))\cdot \cos(\sin(x \cos(x)))\cdot \cos(x\cos x)\cdot (\cos x-x\sin x)\)
\[ f(x) = \arcsin(\arctan(\arccos(x^2 + 1))) \]

Solution:

\(f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\arctan(\arccos(x^2+1))\right)^2}}\cdot(\arctan(\arccos(x^2 + 1)))'\)

\((\arctan(\arccos(x^2 + 1)))'=\frac{1}{1+(\arccos(x^2 + 1))^2}\cdot (\arccos(x^2 + 1))'\)

\((\arccos(x^2+1)^{\prime}=-\frac{1}{\sqrt{1-(x^2+1)^2}}\cdot2x\)

Finally, \(f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\arctan(\arccos(x^2+1))\right)^2}}\cdot\frac{1}{1+(\arccos(x^2 + 1))^2}\cdot -\frac{1}{\sqrt{1-(x^2+1)^2}}\cdot 2x\)

Thus \(f^{\prime}(x)=-\frac{1}{\sqrt{1-\left(\arctan(\arccos(x^2+1))\right)^2}}\cdot\frac{1}{1+(\arccos(x^2 + 1))^2}\cdot \frac{1}{\sqrt{1-(x^2+1)^2}}\cdot 2x\)

3. (25 points) Decide if the following sentences are true or false. Justify properly:

If \(f + g\) is differentiable in \(x = a\), then \(f\) and \(g\) are differentiable in \(x = a\).

False

\(f(x)=\left|x\right|\) and \(g(x)=-|x|\)

We know \((f+g)(x)=0\) is differentiable at \(x=0\), but \(f(x)\) and \(g(x)\) is not differentiable at \(x=0\)
If \(f \cdot g\) is differentiable in \(x = a\), then \(f\) and \(g\) are differentiable in \(x = a\).

False

\(f(x)=\left|x\right|\) and \(g(x)=\left|x\right|\)

We know \((f\cdot g)(x)=x^2\) is differentiable at \(x=0\), but \(f(x)\) and \(g(x)\) is not differentiable at \(x=0\)
If \(f\) is continuous in \(x = a\), then \(|f|\) is continuous in \(x = a\).

If \(f\) is continuous in \(x=a\), then we know \(\forall\varepsilon>0,\exists\delta>0,\left|x-a\right|<\delta,\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon\)

Then we know \(||f(x)|-|f(a)||<|f(x)-f(a)|<\varepsilon\)

Thus \(|f|\) is continuous in \(x=a\)
There is a continuous function in \(\mathbb{R}\) that is not differentiable in an infinite set of points.

True

\(f(x)=\begin{cases}x-2n,x\in\left\lbrack2n,1+2n\right\rbrack\\2\left(n+1\right)-x,x\in\left\lbrack1+2n,2+2n\right\rbrack\end{cases}\) where \(n\in\Z\)

But \(x=1+2n\) is not differentiable

4. (25 points) Suppose that \(f\) and \(g\) are \(n\)-times differentiable functions, then the product \(f g\) is also \(n\)-times differentiable and its \(n\)-th derivative is given by the formula

\[ (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k)}. \]

Proof

Use induction

Basic step: \(\left(fg\right)^{\prime}=f^{\prime}g+fg^{\prime}=\binom10f^{^{\prime}}g^{}+\binom11fg^{\prime}=\sum_{k=0}^1\binom{1}{k}f^{(1-k)}g^{(k)}\)

Inductive step: Suppose it is true for \((fg)^{(n-1)}=\sum_{k=0}^{n-1}\binom{n-1}{k}f^{(n-1-k)}g^{(k)}\)

Then \((fg)^{(n)}=((fg)^{(n-1)})'=(\sum_{k=0}^{n-1}\binom{n-1}{k}f^{(n-1-k)}g^{(k)})'=(\binom{n-1}{0}f^{(n-1)}g)'+(\binom{n-1}{1}f^{(n-2)}g^{(1)})'+...+(\binom{n-1}{n-1}fg^{(n-1)})'\)

\(=\binom{n-1}{0}f^{(n)}g+\left\lbrack\binom{n-1}{0}f^{\left(n-1\right)}g^{\left(1\right)}+\binom{n-1}{1}f^{(n-1)}g^{\left(1\right)}\right\rbrack+\left\lbrack\binom{n-1}{1}f^{\left(n-2\right)}g^{\left(2\right)}+\ldots\right\rbrack+...+\left\lbrack\ldots+\binom{n-1}{n-1}f^{\left(1\right)}g^{(n-1)}\right\rbrack+\binom{n-1}{n-1}fg^{(n)}\)

\(=\binom{n}{0}f^{(n)}g+\left\lbrack\binom{n}{1}f^{\left(n-1\right)}g^{\left(1\right)}\right\rbrack+\left\lbrack\binom{n}{2}f^{\left(n-2\right)}g^{\left(2\right)}\right\rbrack+...+\left\lbrack\binom{n}{n-1}f^{\left(1\right)}g^{(n-1)}\right\rbrack+\binom{n}{n}fg^{(n)}\)

\(=\binom{n}{0}f^{(n)}g+...+\binom{n}{n}fg^{(n)}\)

\(= \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k)}.\)

Thus \((fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k)}\) is true for all \(n\)

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