\(\displaystyle -4\ \textup{and}\ 5\)

\(\displaystyle \left \langle 2e^{10}, 3, 10 \right \rangle\)

\(\displaystyle \frac{3xy}{\sqrt{3x^{2}y+z^{3}}}\)

\(\displaystyle L\left ( x,y \right )=e+3ex+ey\)

Notice that \(\displaystyle f(x)\) can be expressed as a composite function, i.e. a function within a function. If we let \(\displaystyle g(x)=1-x^2\) and  \(\displaystyle h(x)=\ln (x)\), then \(\displaystyle f(x)=h(g(x))\). In order to differentiate \(\displaystyle f(x)\), we will need to apply the Chain Rule, as shown below:

\(\displaystyle f'(x)=h'(g(x))\cdot g'(x)\)

First, we need to find \(\displaystyle h'(x)\), which equals \(\displaystyle \frac{1}{x}\).

Then, we need to find \(\displaystyle g'(x)\) by applying the Product Rule.

\(\displaystyle g'(x)=-2x\)

\(\displaystyle f'(x)=h'(g(x))\cdot g'(x)\)\(\displaystyle =\frac{1}{1-x^2}\cdot (-2x)=\frac{2x}{x^2-1}\)

The answer is \(\displaystyle f'(x)=\frac{2x}{x^2-1}\).

First, we can write out the first few terms of the sequence \(\displaystyle a_{n}=\)\(\displaystyle 4(\frac{-1}{2})^{n-1}\), where \(\displaystyle n\) ranges from 1 to 3.

\(\displaystyle a_{1}=4(\frac{-1}{2})^{1-1}=4\)

\(\displaystyle a_{2}=4(\frac{-1}{2})^{2-1}=-2\)

\(\displaystyle a_{3}=4(\frac{-1}{2})^{3-1}=1\)

Notice that each term \(\displaystyle \left ( n>1 \right )\), is found by multiplying the previous term by \(\displaystyle -\frac{1}{2}\). Therefore, this sequence is a geometric sequence with a common ratio of \(\displaystyle -\frac{1}{2}\). We can find the sum of the terms in an infinite geometric sequence, provided that \(\displaystyle |r|< 1\), where \(\displaystyle r\) is the common ratio between the terms. Because \(\displaystyle r=-\frac{1}{2}\) in this problem, \(\displaystyle \left | r \right |\) is indeed less than 1. Therefore, we can use the following formula to find the sum, \(\displaystyle S\), of an infinite geometric series.

\(\displaystyle S=\frac{a_{1}}{1-r}=\frac{4}{1-(\frac{-1}{2})}=\frac{4}{3/2}=\frac{8}{3}\)

The answer is \(\displaystyle \frac{8}{3}\).

We will have to find the first derivative of \(\displaystyle y\) with respect to \(\displaystyle x\) using implicit differentiation. Then, we can find \(\displaystyle \frac{d^{2}y}{dx^{2}}\), which is the second derivative of \(\displaystyle y\) with respect to \(\displaystyle x\).

\(\displaystyle \frac{d}{dx}[y^2-x^2]=\frac{d}{dx}[4]\)

We will apply the chain rule on the left side.

\(\displaystyle \frac{d}{dx}[y^2]\cdot \frac{dy}{dx}-\frac{d}{dx}[x^2]=0\)

\(\displaystyle 2y\cdot \frac{dy}{dx}-2x=0\)

We now solve for the first derivative with respect to \(\displaystyle x\).

\(\displaystyle \frac{dy}{dx}=\frac{x}{y}\)

In order to get the second derivative, we will differentiate \(\displaystyle \frac{x}{y}\) with respect to \(\displaystyle x\). This will require us to employ the Quotient Rule.

\(\displaystyle \frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}[\frac{x}{y}]=\frac{y\cdot \frac{d}{dx}[x]-x\cdot \frac{d}{dx}[y]}{y^{2}} =\frac{y\cdot 1-x\cdot \frac{dy}{dx}}{y^{2}}\)

We will replace \(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}\) with \(\displaystyle \frac{x}{y}\).

\(\displaystyle \frac{y-x(\frac{x}{y})}{y^{2}}=\frac{y^{2}-x^{2}}{y^{3}}\)

But, from the original equation, \(\displaystyle y^{2}-x^{2}=4\). Also, if we solve for \(\displaystyle y\), we obtain \(\displaystyle y = (4+x^{2})^{\frac{1}{2}}\).

\(\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{y^{2}-x^{2}}{y^{3}}=\frac{4}{(x^{2}+4)^{\frac{3}{2}}}\)

The answer is \(\displaystyle \frac{4}{(x^{2}+4)^{\frac{3}{2}}}\).

\(\displaystyle r = 4\cos\theta\)

First, mulitply both sides by r. 

\(\displaystyle r^2 =4r\cos\theta\)

Then, use the identities \(\displaystyle x^2+y^2 =r^2\) and \(\displaystyle x = r\cos\theta\).

\(\displaystyle x^2+y^2=4x\)

\(\displaystyle x^2-4x+y^2=0\)

\(\displaystyle x^2 -4x + 4 +y^2 =4\)

\(\displaystyle (x-2)^2 +y^2 =4\)

The answer is \(\displaystyle (x-2)^2 +y^2 =4\).

We can use the substitution technique to evaluate this integral.

Let \(\displaystyle u=1-t\).

We will differentiate \(\displaystyle u\) with respect to \(\displaystyle t\).

\(\displaystyle \frac{\mathrm{d}u }{\mathrm{d} t}=-1\), which means that \(\displaystyle {\mathrm{d} u}=-{\mathrm{d} t}\).

We can solve for \(\displaystyle {\mathrm{d}t }\) in terms of \(\displaystyle {\mathrm{d} u}\), which gives us \(\displaystyle {\mathrm{d}t }=-{\mathrm{d} u}\).

We will also need to change the bounds of the integral. When \(\displaystyle t=-1\), \(\displaystyle u=1-(-1)\), and when \(\displaystyle t=0\), \(\displaystyle u=1-0=1\).

We will now substitute \(\displaystyle u\) in for the \(\displaystyle 1-t\), and we will substitute \(\displaystyle -{\mathrm{d}u }\) for \(\displaystyle {\mathrm{d} t}\).

\(\displaystyle \int_{2}^{1}-e^{u}du\)

\(\displaystyle \int_{2}^{1}-e^{u}du = -e^{u}|_{2}^{1}=-e^{1}-(-e^{2})=e^{2}-e^{1}\)

The answer is \(\displaystyle e^{2}-e\).

First, let's multiply the numerator and denominator of the fraction in the limit by \(\displaystyle \frac{1}{x^{4}}\).

\(\displaystyle \lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}\)\(\displaystyle =\lim_{x\rightarrow \infty }\frac{(\frac{1}{x^4})(1-x^4)}{\frac{1}{x^{4}}(x^2-4x^4)}\)

\(\displaystyle =\lim_{x\rightarrow \infty }\frac{\frac{1}{x^4}-1}{\frac{1}{x^2}-4}\)

As \(\displaystyle x\) becomes increasingly large the \(\displaystyle \frac{1}{x^{4}}\) and\(\displaystyle \frac{1}{x^{2}} ^{}\) terms will tend to zero. This leaves us with the limit of \(\displaystyle \frac{1}{4}\).

\(\displaystyle \lim_{x\rightarrow \infty }\frac{-1}{-4}=\frac{1}{4}\).

The answer is \(\displaystyle \frac{1}{4}\).