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Algorithm Homework:
Use the definition of the derivative to show that $\frac{d}{dx}(\cos(x))=-\sin(x)$
Answer
By the definition, we directly compute \(\begin{align} \frac{d}{dx}(\cos(x))& =\lim_{h\to 0}\frac{\cos(x+h)-\cos(x)}{h}=\lim_{h\to 0}\frac{-2\sin\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}\\\\ & =\lim_{h\to 0}\frac{-\sin\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\\\\ & =-\sin(x). \end{align}\)
Find the $n$th derivative of the function $f(x)=x^{n-1}\ln(x)$
Answer
We process by induction to show that the requested derivative is $\frac{(n-1)!}{d}$. Consider $n=1$. The first derivative of $f(x)$ is $\frac{1}{x}$. Suppose the statement is true for some $n$. Let us prove the statement is true for $n+1$. The $n+1$ derivative is the $n$th derivative of the first dierivative of $x^{n}\ln(x)$, so it is the $n$th derivative of $nx^{n-1}+x^{n-1}$. The $n$th derivative of $x^{n-1}\ln(x)$ is $\frac{(n-1)!}{x}$ by the induction hypothesis, so we can conclude that the $n+1$th derivative is $\frac{n!}{x}$.
Calculate $y’$. $y=\frac{x^2-x+2}{\sqrt{x}}$
Answer
We simplify $y=\frac{x^2-x+2}{\sqrt{x}} = x^{3/2}-x^{1/2}+2x^{-1/2}$, so $y’ = \frac{3}{2}x^{1/2}-\frac{1}{2}x^{-1/2}-x^{-3/2}$.
Calculate $y’$. $y=\frac{\tan x}{1+\cos x}$
Answer
Applying the quotient rule, we have $y’ = \frac{\sec^2x(1+\cos x) +\tan(x)\sin(x)}{(1+\cos x)^2}$.
Answer
Applying the chain rule and the multiplication rule, we get $y’ = \frac{1}{x\ln(x)}(\ln(x)+1)$.
Calculate $y’$. $y=e^{x\sec x}$
Answer
Applying the chain rule and the multiplication rule, we get $y’ = e^{x\sec x}(\sec x+x\sec x\tan x)$.
Calculate $y’$. $\sin(xy)=x^2-y$
Answer
Applying the implicit differentiation, we have $\cos(xy)(y+xy’) = 2x - y’$, so $y’ =\frac{2x-\cos(xy)y}{x+1}$.
| Calculate $y’$. $y=\ln | \sec 5x+\tan 5x | $ |
Answer
Applying the chain rule, we have $y’ = \frac{1}{\sec 5x +\tan 5x}(5\sec 5x\tan 5x+5\sec^2 5x)$.
If $f(x)=(x-a)(x-b)(x-c)$, show that \(\frac{f'(x)}{f(x)}=\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c}\)
Answer
We have $f’(x) = (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b)$ by the general multiplication rule, so we can attain the conclusion by trivial algebraic manipulation.
Suppose that $f(x)$ and $g(x)$ are differentiable on an open interval containing $[a,b]$ with $f(a)>g(a)$ and $f(b)>g(b)$. If $f(x)=g(x)$ has exactly one solution on $[a,b]$, denoted by $x_0\in [a,b]$, then $f$ and $g$ have the same tangent line at $x=x_0$.
Answer
There are three known arguments for this question. Here, I will provide one of them. Let $h(x)=f(x)-g(x)$. Then, $f(x)=g(x)$ has exactly one solution on $[a,b]$ if and only if $h(x)=0$ has exactly one solution on $[a,b]$. It implies $h(x_0)=0$ is a global minimum of $h$. If it is not, then we have some $x_1$ such that $f(x_1)<0$. Then, by the intermediate value theorem, $h(x)=0$ have a solution between $[a,x_1)$ and another solution between $(x_1,b]$, which is a contradiction. Therefore, $h(x_0)$ is a global minimum. By Fermat’s theorem, we then have $h’(x_0)=f’(x_0)-g’(x_0)=0$, wich derive the desired conclustion.
- State the intermediate value theorem
- Develop a pseudocode for approximating a solution of $\sqrt{x^2-5}-3=0$ with an error less than $10^{-8}$ utilizing the Intermediate Value Theorem and the Bisection Method. Elaborate on the importance of selecting the initial endpoints carefully.
- Implement the pseudocode in Sage/Python.
Answer
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