Take away the parentheses by multiplying the binomial by itself so that you are left with a polynomial. 

\(\displaystyle f(x)=(3x+1)^2=9x^2+6x+1\)

Then take the derivative of each term in the polynomial using the power rule \(\displaystyle nx^{n-1}\) where n is the exponent:

\(\displaystyle \\f'(x)=\frac{d}{dx}(9x^2+6x+1)\\ \\=\frac{d}{dx}(9x^2)+\frac{d}{dx}(6x)+\frac{d}{dx}(1)\\ \\=2\cdot 9x^{2-1}+1\cdot 6x^{^{1-1}}\\ \\=18x+6\)

Take away the parentheses by multiplying the outside piece to each term within the binomial so that you are left with a polynomial. 

\(\displaystyle f(x)=\sqrt{x}(x-1)=x^{\frac{3}{2}}-x^{\frac{1}{2}}\)

Then take the derivative of each term in the polynomial using the power rule \(\displaystyle nx^{n-1}\) where n is the exponent:

\(\displaystyle \\f'(x)=\frac{d}{dx}(x^{\frac{3}{2}}-x^{\frac{1}{2}})\\ \\=\frac{3}{2}x^{\frac{3}{2}-1}-\frac{1}{2}x^{\frac{1}{2}-1}\\ \\=\frac{3}{2}x^{\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\)

Bring the root to the numerator. Take away the parentheses by multiplying the outside piece by each of the terms in the parentheses so that you are left with a polynomial. 

\(\displaystyle f(x)=\frac{x^2+3x+4}{\sqrt{x}}=(\sqrt{x})^-^1(x^2+3x+4)=x^{\frac{3}{2}}+3x^{\frac{1}{2}}+4x^{-\frac{1}{2}}\)

Then take the derivative of each term in the polynomial using the power rule \(\displaystyle nx^{n-1}\) where n is the exponent:

\(\displaystyle \\ f'(x)=\frac{d}{dx}(x^{\frac{3}{2}}+3x^{\frac{1}{2}}+4x^{-\frac{1}{2}})\\ \\=\frac{3}{2}x^{\frac{3}{2}-1}+\frac{1}{2}\cdot 3x^{\frac{1}{2}-1}+\left(-\frac{1}{2}\cdot 4x^{-\frac{1}{2}-1}\right)\\ \\=\frac{3}{2}x^{\frac{1}{2}}+\frac{3}{2}x^{-\frac{1}{2}}-2x^{-\frac{3}{2}}\)

Take the derivative of each term in the polynomial using the power rule \(\displaystyle nx^{n-1}\) where n is the exponent:

\(\displaystyle f(x)=x\sqrt{3}+\sqrt{2x}=x\sqrt{3}+x^{\frac{1}{2}}\sqrt{2}\)

\(\displaystyle \\f'(x)=\frac{d}{dx}(x\sqrt{3}+x^{\frac{1}{2}}\sqrt{2})\\ \\=1\cdot x^{1-1}\cdot \sqrt{3}+\frac{1}{2}x^{\frac{1}{2}-1}\cdot \sqrt{2}\\ \\=\sqrt{3}+\frac{1}{2}\cdot \sqrt{2}x^{-\frac{1}{2}}\\ \\=\sqrt{3}+x^{-\frac{1}{2}}\cdot \frac{\sqrt{2}}{2}\)

Take away the parentheses by multiplying the outside piece by each term within the parentheses so that you are left with a polynomial. 

\(\displaystyle f(x)=3x(\frac{1}{x}+x^2)=3+3x^3\)

Then take the derivative of each term in the polynomial using the power rule \(\displaystyle nx^{n-1}\) where n is the exponent:

\(\displaystyle \\f'(x)=\frac{d}{dx}(3+3x^3)\\ \\=\frac{d}{dx}(3)+\frac{d}{dx}(3x^3)\\ \\=0+3\cdot x^{3-1}\\ \\=9x^2\)

To solve this problem, we must first know that the surface area of a box with a square bottom would be \(\displaystyle A=4sh+s^2\) where \(\displaystyle s\) is the length of the side of the box and \(\displaystyle h\) is the height of the box.

The volume of the box can be found with the formula \(\displaystyle V=s^2h\) where \(\displaystyle s\) is the length of the side of the box and \(\displaystyle h\) is the height of the box.

We know that the volume of such a box must be equivalent to 6,250; therefore to solve this problem we must first solve one of the variables in terms of another.

\(\displaystyle \frac{6,250}{s^2}=h\)

Then we must plug it into the surface area equation.

\(\displaystyle A=\frac{(4)(6,250)}{s}+s^2\)

Then, take the derivative and set it equal to zero to find the minimum. To take the derivative of that problem, one must know the power rule 

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\).

\(\displaystyle 0=-\frac{(4)(6,250)}{s^2}+2s\)

Solving for \(\displaystyle s\)

\(\displaystyle 2s^3=25,000\)

\(\displaystyle s^3=12,500\)

\(\displaystyle s=23.2 mm\)

Plugging back into the original volume equation to solve for \(\displaystyle h\).

\(\displaystyle 6,250=(23.2)^2h\)

\(\displaystyle h=11.61mm\)

To approach this problem you msut first understand the different rules of calculus.  

First is the Quotient rule, 

\(\displaystyle \frac{d}{dx}\frac{u}{v}=\frac{u'v-uv'}{v^2}\).  

Second is the product rule, 

\(\displaystyle \frac{d}{dx}uv=u'v+uv'\).  

Third is the power rule, 

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\).  

Fourth is the chain rule,

\(\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))g'(x)\).  

Lastly, some identities must be known these are \(\displaystyle \frac{d}{dx}sinx=cosx\) and \(\displaystyle \frac{d}{dx}ln(x)=\frac{1}{x}\).

The step by step solution of the equation above is as follows

First we apply the quotient rule and power rule.

\(\displaystyle y'=\frac{x^4(ln(x)sin(2x))'-4x^3ln(x)sin(2x)}{x^8}\)

Next the product rule and chain rule as well as the identities to take the derivative of \(\displaystyle ln(x)sin(2x)\), giving our final answer.

\(\displaystyle y'=\frac{x^4(ln(x)2cos(2x)+\frac{sin(2x)}{x})-4x^3ln(x)sin(2x)}{x^8}\)

The first few derivatives can be evaluated to give a general trend:

\(\displaystyle y=sin(x)+e^{2x}\)

\(\displaystyle y'=cos(x)+2e^{2x}\)

\(\displaystyle y''=-sin(x)+4e^{2x}\)

\(\displaystyle y^{(3)}=-cos(x)+8e^{2x}\)

\(\displaystyle y^{(4)}=cos(x)+16e^{2x}\)

Thus:

\(\displaystyle n=1,5,9,13,..., y^{(n)}=cos(x)+2^ne^{2x}\)

\(\displaystyle n=2,6,10,14,..., y^{(n)}=-sin(x)+2^ne^{2x}\)

\(\displaystyle n=3,7,11,15,..., y^{(n)}=-cos(x)+2^ne^{2x}\)

\(\displaystyle n=4,8,12,16,..., y^{(n)}=sin(x)+2^ne^{2x}\)

Therefore when \(\displaystyle n=11\):

\(\displaystyle y^{(11)}=-cos(x)+2^{11}e^{2x}\)

To differentiate this function, you must use the chain rule. To do that, you first multiply the exponent by the coefficient in front of the expression. In this case, it's one. Then, you subtract one from the exponent, so at this point, you have \(\displaystyle 6(3x^2+4x)^5\).

The last step is to take the derivative of what's inside the parantheses, which is \(\displaystyle 6x+4\).

Multiply like terms so that your derivative is

\(\displaystyle f'(x)=(36x+24)(3x^2+4x)^5\)

It might be helpful to rewrite this equation first: \(\displaystyle y=(1+x^2)^{\frac{1}{2}}\). Then, apply the chain rule.

First, multiply the exponent by the coefficient in front of the expression (in this case, 1).

Then, subtract 1 from the exponent.

Lastly, take the derivative of the expression.

This comes out to

\(\displaystyle y=\left(\frac{1}{2}\right)(1+x^2)^{\frac{-1}{2}}(2x)\).

Then, multiply like terms and get rid of the negative exponent:

\(\displaystyle y=\frac{x}{\sqrt{1+x^2}}\).