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这是惠灵顿维多利亚大学微积分课程的代写成功案例。
MATH141课程简介
This course provides a thorough development of the differential calculus and an introduction to the integral calculus. The course builds on the ideas of functions and limits to define derivatives and integrals, as well as rules for computing these and applications to physical modelling.
This course is designed for in-person study, and it is strongly recommended that students attend lectures and tutorials on campus. In particular some assessment items will have a requirement of in-person attendance, although exceptions can be made under special circumstances.
Queries about any such exceptions can be sent to [email protected].
Prerequisites
This course provides a thorough introduction to the differential calculus and an introduction to the integral calculus. Following a review of coordinate geometry and the use of equations to represent straight lines and circles, we introduce the concept of a function and many important examples including polynomial, rational, trigonometric, exponential and logarithm functions. The idea of a limit is central and leads to considering continuity and differentiability of a function. The definition of a derivative and rules for computing derviatives are deduced. Derivatives are applied to model physical problems. The course concludes by introducing the idea of an integral of a function and we demonstrate the fundamental property that integrals can be calculated using the inverse of differentiation.As well as developing methods of computation and applications of the calculus, the course focuses on its underlying concepts, the correct use of symbolic representation in mathematics and the importance of providing logical justification for its results and methods.
MATH141 Calculus HELP（EXAM HELP， ONLINE TUTOR）
Use the definition of the derivative to compute $f^{\prime}\left(x_0\right)$ in each of the following cases.
$$
f(x)=(3 x+1)^2, \quad f(x)=\left(x^2-1\right)^2 .
$$
The derivative of the first function is given by the limit
$$
\begin{aligned}
f^{\prime}\left(x_0\right) & =\lim {x \rightarrow x_0} \frac{(3 x+1)^2-\left(3 x_0+1\right)^2}{x-x_0}=\lim {x \rightarrow x_0} \frac{\left(3 x-3 x_0\right)\left(3 x+3 x_0+2\right)}{x-x_0} \
& =\lim _{x \rightarrow x_0} 3\left(3 x+3 x_0+2\right)=3\left(6 x_0+2\right)=6\left(3 x_0+1\right)
\end{aligned}
$$
The derivative of the second function is given by the limit
$$
\begin{aligned}
f^{\prime}\left(x_0\right) & =\lim {x \rightarrow x_0} \frac{\left(x^2-1\right)^2-\left(x_0^2-1\right)^2}{x-x_0}=\lim {x \rightarrow x_0} \frac{\left(x^2-x_0^2\right)\left(x^2+x_0^2-2\right)}{x-x_0} \
& =\lim _{x \rightarrow x_0}\left(x+x_0\right)\left(x^2+x_0^2-2\right)=2 x_0\left(2 x_0^2-2\right)=4 x_0\left(x_0^2-1\right) .
\end{aligned}
$$
Compute the derivative $y^{\prime}=\frac{d y}{d x}$ in each of the following cases.
$$
y=\ln (\tan x)+2(\sec x)^5, \quad y=\tan ^{-1}(\sin (2 x))
$$
When it comes to the first function, one may use the chain rule to get
$$
\begin{aligned}
y^{\prime} & =\frac{1}{\tan x} \cdot(\tan x)^{\prime}+10(\sec x)^4 \cdot(\sec x)^{\prime} \
& =\frac{1}{\tan x} \cdot \sec ^2 x+10 \sec ^4 x \cdot \sec x \tan x=\frac{\sec ^2 x}{\tan x}+10 \sec ^5 x \cdot \tan x .
\end{aligned}
$$
When it comes to the second function, one similarly finds that
$$
y^{\prime}=\frac{1}{\sin ^2(2 x)+1} \cdot \sin (2 x)^{\prime}=\frac{2 \cos (2 x)}{\sin ^2(2 x)+1} .
$$
Compute the derivative $f^{\prime}\left(x_0\right)$ in the case that
$$
f(x)=\frac{\left(x^3+2\right)^3 \cdot e^{4 x} \cdot \cos (5 \tan x)}{\sqrt{x^3+1}}, \quad x_0=0 .
$$
First, we use logarithmic differentiation to determine $f^{\prime}(x)$. In this case, we have
$$
\begin{aligned}
\ln |f(x)| & =\ln \left|x^3+2\right|^3+\ln e^{4 x}+\ln |\cos (5 \tan x)|+\ln \left|x^3+1\right|^{-1 / 2} \
& =3 \ln \left|x^3+2\right|+4 x+\ln |\cos (5 \tan x)|-\frac{1}{2} \ln \left|x^3+1\right| .
\end{aligned}
$$
Differentiating both sides of this equation, one easily finds that
$$
\frac{f^{\prime}(x)}{f(x)}=\frac{3 \cdot 3 x^2}{x^3+2}+4-\frac{\sin (5 \tan x) \cdot 5 \sec ^2 x}{\cos (5 \tan x)}-\frac{3 x^2}{2\left(x^3+1\right)} .
$$
To compute the derivative $f^{\prime}(0)$, one may then substitute $x=0$ to conclude that
$$
\frac{f^{\prime}(0)}{f(0)}=0+4-0-0=4 \quad \Longrightarrow \quad f^{\prime}(0)=4 f(0)=32
$$
Show that the derivative of the inverse tangent function is given by
$$
\left(\tan ^{-1} x\right)^{\prime}=\frac{1}{1+x^2} .
$$
Using Theorem 3.19 with $f(x)=\tan x$ and $g(x)=\tan ^{-1} x$, one finds that
$$
g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}=\frac{1}{\sec ^2(g(x))}=\cos ^2 g(x)=\cos ^2\left(\tan ^{-1} x\right)
$$
Let $\theta=\tan ^{-1} x$ for simplicity and note that $\tan \theta=x$. When $x \geq 0$, the angle $\theta$ arises in a right triangle with an opposite side of length $x$ and an adjacent side of length 1 . It follows by Pythagoras’ theorem that the hypotenuse has length $\sqrt{1+x^2}$, so
$$
g^{\prime}(x)=\cos ^2\left(\tan ^{-1} x\right)=\cos ^2 \theta=\left(\frac{1}{\sqrt{1+x^2}}\right)^2=\frac{1}{1+x^2} .
$$
When $x \leq 0$, the last equation holds with $-x$ instead of $x$. This changes the term $\tan ^{-1} x$ by a minus sign, but the cosine remains unchanged, so the equation is still valid.
MY-ASSIGNMENTEXPERT™可以为您提供WGTN.AC.NZ MATH141 CALCULUS微积分的代写代考和辅导服务！
