To solve this, first find the derivative of the function (otherwise known as the slope). 

y = ln(x²)

y' = (2x/(x²))

Then, to find the slope in respect to the given points (e, 3), plug in e. 

y' = (2e)/(e²)

Simplify. 

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3). 

y – 3 = (2/e)(x – e)

The answer is 

\(\displaystyle y=\frac{x^{3}-2x(x^{1/2})}{x}\)    let's go ahead and cancel out the \(\displaystyle x\)'s. This will simplify things.

\(\displaystyle y=x^{2}-2(x^{1/2})\)

\(\displaystyle y'=2x-\frac{1}{(x^{1/2})}\)

\(\displaystyle y'(1)=2(1)-\frac{1}{(1^{1/2})} =1\) this is the slope so let's use the point slope formula.

\(\displaystyle y+1=1(x-1)\)

\(\displaystyle y+1=x-1\)

\(\displaystyle y=x-2\)

We see the answer is \(\displaystyle y'=\frac{-1}{(t-6)^{2}}\) after we simplify and use the quotient rule.

\(\displaystyle y=\frac{t+2}{t^{2}-4t-12}\) we could use the quotient rule immediatly but it is easier if we simplify first.

\(\displaystyle y=\frac{t+2}{(t+2)(t-6)}\)

\(\displaystyle y=\frac{1}{(t-6)}\)

\(\displaystyle y=(t-6)^{-1}\)

\(\displaystyle y'=-(t-6)^{-2}\)

\(\displaystyle y'=\frac{-1}{(t-6)^{2}}\)

When taking limits to infinity, we usually only consider the highest exponents. In this case, the numerator has \(\displaystyle -2x^3\) and the denominator has \(\displaystyle x^2\). Therefore, by cancellation, it becomes \(\displaystyle -2x\) as \(\displaystyle x\) approaches infinity. So the answer is \(\displaystyle -\infty\).

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 \(\displaystyle \frac{8}{3}\).

The answer is 1.

\(\displaystyle h(x)=\frac{2f}{g}\)

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

\(\displaystyle h'(x)=\frac{2(3*2-4*1)}{4} = 1\)

The answer is \(\displaystyle 10x-39\)

\(\displaystyle f(x)=(x-2)(x+4)\)

\(\displaystyle f(x)=x^{2}+2x-8\)

\(\displaystyle f'(x)=2x+2\)

\(\displaystyle f'(x) = 2(4)+2=10\) 

(This is the slope. Now use the point-slope formula)

\(\displaystyle y-1=10(x-4)\)

\(\displaystyle y-1=10x-40\)

\(\displaystyle y=10x-39\)

The answer is \(\displaystyle y=2ax+bx-2a-b+1\)

\(\displaystyle f(x)=ax^{2}+bx+c\)

\(\displaystyle f'(x)=2ax+b\)

\(\displaystyle f'(1)=2a(1)+b = 2a+b\) 

(This is the slope. Now use the point-slope formula.)

\(\displaystyle y-1=(2a+b)(x-1)\)

\(\displaystyle y-1=2ax+bx-2a-b\)

\(\displaystyle y=2ax+bx-2a-b+1\)

The answer is \(\displaystyle -1\).

\(\displaystyle f(x)=cos(x)\)

\(\displaystyle f'(x)=-sin(x)\) 

We know that

\(\displaystyle \frac{\pi }{2}=90^{\circ}\)

so,

\(\displaystyle f'\left ( \frac{\pi }{2}\right )=-sin(90^{\circ})=-1\)

When we let x = 0 in our original limit, we obtain the 0/0 indeterminate form. Therefore, we can apply L'Hospital's Rule, which requires that we take the derivative of the numerator and denominator separately.

\(\displaystyle \lim_{x\rightarrow 0}\frac{2\sin(x^2)}{x\tan(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}2\sin(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}x\tan(x)}\)

Apply the Chain Rule in the numerator and the Product Rule in the denominator.

\(\displaystyle \lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}\)

If we again substitute x = 0, we still obtain the 0/0 indeterminate form. Thus, we can apply L'Hospital's Rule one more time.

\(\displaystyle \lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}4x\cos(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}(\tan(x)+x\sec^2(x))}\)

\(\displaystyle =\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}\)

If we now let x = 0, we can evaluate the limit.

\(\displaystyle =\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}=\frac{4\cos(0)+0}{\sec^2(0)+\sec^2(0)+0}\)

\(\displaystyle =\frac{4}{1+1}=2\)

The answer is 2.