Calculate the dot product of the 2 vectors.

In general,

\(\displaystyle \vec{v} = (v_1,v_2,v_3)\)

\(\displaystyle \vec{u} = (u_1,u_2,u_3)\)

\(\displaystyle \vec{v} \cdot \vec{u} = v_1 * u_1 + v_2 * u_2 + v_3 * u_3\)

Solution:

\(\displaystyle \vec{v} =((3x^2+3),e^{2x},(sin(x)+2x))\)

\(\displaystyle \vec{u} =(2x,(4x+2),-1)\)

\(\displaystyle \vec{v}\cdot\vec{u}=((3x^2+3),e^{2x},(sin(x)+2x)\cdot(2x,(4x+2),-1)\)

\(\displaystyle \vec{v}\cdot\vec{u}=(3x^2+3)*2x+e^{2x}*(4x+2)+(sin(x)+2x)*-1\)

\(\displaystyle \vec{v}\cdot\vec{u}=6x^3+6x+4xe^{2x}+2e^{2x}+(-sin(x)-2x)\)

\(\displaystyle \vec{v}\cdot\vec{u}=6x^3+4x+4xe^{2x}+2e^{2x}-sin(x)\)

Calculate the sum of vectors.

In general,

\(\displaystyle \vec{v} = (v_1,v_2,v_3)\)

\(\displaystyle \vec{u} = (u_1,u_2,u_3)\)

\(\displaystyle \vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)\)

Solution:

\(\displaystyle \vec{v} =(x^2+2x,5x^3-2x+1,2x+6)\)

\(\displaystyle \vec{u} =(2x^2+2x+4,x^2+x,x^2+x)\)

\(\displaystyle \vec{v}+\vec{u}=(x^2+2x,5x^3-2x+1,2x+6)+(2x^2+2x+4,x^2+x,x^2+x)\)

\(\displaystyle \vec{v}+\vec{u}=(x^2+2x+2x^2+2x+4,5x^3-2x+1+x^2+x,2x+6+x^2+x)\)

\(\displaystyle \vec{v}+\vec{u}=(3x^2+4x+4,5x^3+x^2-x+1,x^2+3x+6)\)

Calculate the dot product of the 2 vectors.

In general,

\(\displaystyle \vec{v} = (v_1,v_2,v_3)\)

\(\displaystyle \vec{u} = (u_1,u_2,u_3)\)

\(\displaystyle \vec{v} \cdot \vec{u} = v_1 * u_1 + v_2 * u_2 + v_3 * u_3\)

Solution:

\(\displaystyle \vec{v}\cdot\vec{u}=(5x,16+x,2x^2)\cdot(e^x,2,3x)\)

\(\displaystyle \vec{v}\cdot\vec{u}=5x*e^x+(16+x)*2+2x^2*3x\)

\(\displaystyle \vec{v}\cdot\vec{u}=5xe^x+32+2x+6x^3\)

Calculate the dot product of the 2 vectors.

In general,

\(\displaystyle \vec{v} = (v_1,v_2,v_3)\)

\(\displaystyle \vec{u} = (u_1,u_2,u_3)\)

\(\displaystyle \vec{v} \cdot \vec{u} = v_1 * u_1 + v_2 * u_2 + v_3 * u_3\)

Solution:

\(\displaystyle \newline\vec{v} =\left(\left(\frac{\sqrt{2}+1}{2}\right),\frac{2x}{e^{x}},-2\right)\newline\vec{u} =\left(\left(4+x\right),ln(x),2\right)\newline\)

\(\displaystyle \vec{v}\cdot\vec{u}=((\frac{\sqrt{2}+1}{2}),\frac{2x}{e^{x}},-2)\cdot((4+x),ln(x),2)\)

\(\displaystyle \vec{v}\cdot\vec{u}=(\frac{\sqrt{2}+1}{2})*(4+x)+\frac{2x}{e^{x}}*ln(x)+-2*2\)

\(\displaystyle \vec{v}\cdot\vec{u}=2\sqrt{2}+2 + (\frac{\sqrt{2}+1}{2})*x+\frac{2xln(x)}{e^{x}}+-4\)

\(\displaystyle \vec{v}\cdot\vec{u}=2\sqrt{2}-2+\frac{(\sqrt2+1)}{2}x+\frac{2xln(x)}{e^x}\)

For the two vectors 

\(\displaystyle a=\left< a_1,a_2\right>\) and \(\displaystyle b=\left< b_1,b_2\right>\)

The sum of the two vectors a+b is 

\(\displaystyle \left< a_1, a_2\right> + \left< b_1, b_2\right>= \left< a_1+b_1, a_2+b_2\right>\)

For the vectors u and v in this problem the sum u+v is 

\(\displaystyle \left< 3, 7\right> + \left< 7, 10\right>\)

\(\displaystyle = \left< 3+7, 7+10\right>\)

\(\displaystyle =\left < 10, 17\right>\)

Evaluating the limit directly will produce an indeterminant solution of \(\displaystyle \small \tiny{\frac{0}{0}}\).

The limit definition of a derivative is \(\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)\). However, the alternative form, \(\displaystyle \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(x)\), better suits the given limit.

Let \(\displaystyle f(x)=\sin^2x\) and notice \(\displaystyle f\left(-\frac{\pi}{2}\right)=1\). It follows that \(\displaystyle {\lim_{x \rightarrow -\frac{\pi}{2}} \frac{\sin^2-1}{x-\left(-\frac{\pi}{2} \right )} = \frac{d}{dx}\sin^2x}\).  

Thus, the limit is \(\displaystyle {2\sin\left(-\frac{\pi}{2} \right ) \cos\left(-\frac{\pi}{2}\right)} = 0.\)

Evaluating the derivative directly will produce an indeterminant solution of \(\displaystyle \small \tiny{\frac{0}{0}}\).

The limit definition of a derivative is \(\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)\). However, the alternative form, \(\displaystyle \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(x)\), better suits the given limit.

Let \(\displaystyle f(x)=x^{49}\) and notice \(\displaystyle f(1)=1\). It follows that \(\displaystyle {\lim_{x \rightarrow 1} \frac{x^{49}-1}{x-1} = \frac{d}{dx}x^{49}}\).  Thus, the limit is \(\displaystyle 49(1)^{48}=49\).

The correct answer is 11.

Taking the derivative of \(\displaystyle h(x)\) involves the product rule, and the chain rule.

\(\displaystyle h'(x)=[f(x+2)][-g'(-x)]+[g(-x)][f'(x+2)]\)

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

Substituting \(\displaystyle x=1\) into both sides of the derivative we get

\(\displaystyle h'(1)= -g'(-1)f(3)+g(-1)f'(3) = -1\cdot1 + 3 \cdot 4 = 11\).

If we recall the definition of a derivative of a function \(\displaystyle \small f(x)\) at a point \(\displaystyle \small x\), one of the definitions is 

\(\displaystyle \small f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\).

If we compare this definition to the limit 

\(\displaystyle \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}\)

we see that that this is the limit definition of a derivative, so we need to find the function \(\displaystyle \small f(x)\) and the point \(\displaystyle \small x\) at which we are evaluating the derivative at. It is easy to see that the function is \(\displaystyle \small f(x)=\tan x\) and the point is \(\displaystyle \small x=0\). So finding the limit above is equivalent to finding \(\displaystyle \small f'(0)\).

We know that the derivative is \(\displaystyle \small f'(x)=\sec^2(x)\), so we have

\(\displaystyle \small \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}=f'(0)=\sec^2(0)=1^2=1\).

Write the definition of the limit.

\(\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\)

Substitute \(\displaystyle x=2\).

\(\displaystyle \lim_{h\to 0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt2}{h}\)

Since \(\displaystyle h\) is approaching to zero, it would be best to evaluate when we assume that \(\displaystyle h\) is progressively decreasing.  Let's assume \(\displaystyle h=0.1, 0.0001, 0.000001\) and check the pattern.

\(\displaystyle \frac{\sqrt{2+0.1}-\sqrt2}{0.1}=0.349241\)

\(\displaystyle \frac{\sqrt{2+0.0001}-\sqrt2}{0.0001}=0.353549\)

\(\displaystyle \frac{\sqrt{2+0.000001}-\sqrt2}{0.000001}=0.353553\)

The best answer is:  \(\displaystyle 0.35355\)