# 微积分代写 Calculus|MATH 19 Stanford University Assignment

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## Instructions:

This course covers the basics of differential calculus, which is a branch of calculus that deals with rates of change and slopes of curves. The course begins with a review of elementary functions, including exponentials and logarithms, which are essential tools in calculus.

The course then covers the concept of limits, which is a fundamental concept in calculus. Limits allow us to describe the behavior of functions near certain points, and they provide the foundation for the definition of the derivative.

The course then introduces the derivative and its properties, including the power rule, product rule, quotient rule, and chain rule. The derivative measures the instantaneous rate of change of a function at a given point, and it has many important applications in fields such as physics, engineering, and economics.

The course also covers applications of the derivative, including optimization problems, related rates problems, and curve sketching. These topics are important in many areas of science, engineering, and economics.

Prerequisites for the course include trigonometry, advanced algebra, and analysis of elementary functions, including exponentials and logarithms. Students must also take the math placement diagnostic offered through the Math Department website in order to register for the course. This diagnostic test helps ensure that students have the necessary background knowledge and skills to succeed in the course.

Let f(x) = 5 sin x + 3 cos x. Then f(117)(π) =___ .

To find the 117th derivative of $f(x) = 5 \sin x + 3 \cos x$ evaluated at $x = \pi$, we can use the fact that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using this, we can see that:

\begin{align*} f'(x) &= 5\cos x – 3\sin x \ f”(x) &= -5\sin x – 3\cos x \ f”'(x) &= -5\cos x + 3\sin x \ f^{(4)}(x) &= 5\sin x – 3\cos x \ f^{(5)}(x) &= 5\cos x + 3\sin x \ f^{(6)}(x) &= -5\sin x + 3\cos x \ \end{align*}

We can see that the derivatives of $f(x)$ have a repeating pattern with a period of four, so we can use this to find the 117th derivative of $f(x)$:

\begin{align*} f^{(117)}(x) &= f^{(4\cdot29+1)}(x) \ &= f^{(1)}(x) \ &= 5\cos x – 3\sin x \ \end{align*}

Therefore, evaluating this expression at $x = \pi$, we get:

\begin{align*} f^{(117)}(\pi) &= 5\cos \pi – 3\sin \pi \ &= 5(-1) – 3(0) \ &= -5 \ \end{align*}

So, $f^{(117)}(\pi) = -5$.

Let f(x) = 4 cos x − 7 sin x. Then f(87)(0) =____ .

To find the 87th derivative of $f(x) = 4 \cos x – 7 \sin x$ evaluated at $x = 0$, we can use the fact that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using this, we can see that:

\begin{align*} f'(x) &= -4\sin x – 7\cos x \ f”(x) &= -4\cos x + 7\sin x \ f”'(x) &= 4\sin x + 7\cos x \ f^{(4)}(x) &= 4\cos x – 7\sin x \ f^{(5)}(x) &= -4\sin x – 7\cos x \ f^{(6)}(x) &= -4\cos x + 7\sin x \ \end{align*}

We can see that the derivatives of $f(x)$ have a repeating pattern with a period of four, so we can use this to find the 87th derivative of $f(x)$:

\begin{align*} f^{(87)}(x) &= f^{(4\cdot21+3)}(x) \ &= f^{(3)}(x) \ &= 4\sin x + 7\cos x \ \end{align*}

Therefore, evaluating this expression at $x = 0$, we get:

\begin{align*} f^{(87)}(0) &= 4\sin 0 + 7\cos 0 \ &= 0 + 7(1) \ &= 7 \ \end{align*}

So, $f^{(87)}(0) = 7$.

For the circle $x^2+y^2-1=0$ use implicit differentiation to show that $y^{\prime \prime}=-\frac{1}{y^3}$ and $y^{\prime \prime \prime}=-\frac{3 x}{y^5}$.

Starting with the given equation $x^2+y^2-1=0$, we take the derivative with respect to $x$ of both sides, using the chain rule for the $y$-term:

$$\frac{d}{dx}(x^2+y^2-1) = \frac{d}{dx}(0)$$

Using the power rule and chain rule, we get:

$$2x + 2y \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$, we have:

$$\frac{dy}{dx} = -\frac{x}{y}$$

Taking the derivative of this expression with respect to $x$ using the quotient rule, we get:

\begin{align*} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(-\frac{x}{y}\right)\ &= -\frac{y\frac{d}{dx}(x)-x\frac{d}{dx}(y)}{y^2}\ &= -\frac{y(1\cdot y + x\cdot y’) – x(0\cdot y + 1\cdot y’)}{y^2} \ &= -\frac{y^{\prime}\cdot x}{y^2} – \frac{1}{y} \end{align*}

Simplifying this expression, we get:

$$y^{\prime \prime} = -\frac{1}{y^3}$$

Taking the derivative of $y^{\prime}$ with respect to $x$ using the quotient rule, we have:

\begin{align*} \frac{d^3y}{dx^3} &= \frac{d}{dx}\left(-\frac{y^{\prime}}{y}\right)\ &= -\frac{y\frac{d}{dx}(y’)-y’\frac{d}{dx}(y)}{y^2}\ &= -\frac{y^{\prime\prime}\cdot y – y^{\prime}\cdot y’}{y^2}\ &= -\frac{y^{\prime\prime}\cdot y}{y^2} + \frac{y^{\prime}\cdot y’}{y^2} \end{align*}

Substituting the values of $y^{\prime\prime}$ and $y^{\prime}$ obtained earlier, we have:

\begin{align*} y^{\prime \prime \prime} &= -\frac{y^{\prime\prime}\cdot y}{y^2} + \frac{y^{\prime}\cdot y’}{y^2}\ &= -\frac{(-1/y^3) \cdot y}{y^2} + \frac{(-x/y^2)\cdot(-x/y)}{y^2}\ &= \frac{x}{y^5} – \frac{1}{y^3}\ &= -\frac{3x}{y^5} \ \ \ \ \ \text{(using } x^2+y^2=1 \text{)} \end{align*}

Therefore, $y^{\prime \prime}=-\frac{1}{y^3}$ and $y^{\prime \prime \prime}=-\frac{3 x}{y^5}$ for the circle $x^2+y^2-1=0$