This problem requires the product rule. The derivative using the product rule is as follows:

\(\displaystyle \frac{dy}{dx}= f(x)g'(x)+g(x)f'(x)\)

Let \(\displaystyle f(x)=x^5\) and \(\displaystyle g(x)= \ln x\).

Their derivatives are:

\(\displaystyle f'(x)=5x^4\) and \(\displaystyle g'(x)=\frac{1}{x}\)

Substitute the functions into the product rule formula and simplify.

\(\displaystyle (x^5)\left(\frac{1}{x}\right)+(\ln x)(5x^4)=x^4+5x^4 \ln x\)

Recall the chain rule from calculus:

\(\displaystyle (fg)'=f'g+fg'\)

So we will want to begin by finding the first derivative of each of our functions:

\(\displaystyle f(x)=4x^2-5x\)

\(\displaystyle f'(x)=8x-5\)

\(\displaystyle g(x)=-5x^3+6x^2-7\)

\(\displaystyle g'(x)=-15x^2+12x\)

Next, use the chain rule formula:

\(\displaystyle (fg)'=(8x-5)(-5x^3+6x^2-7)+(4x^2-5x)(-15x^2+12x)\)

Expand everything to get

\(\displaystyle (fg)'=(-40x^4+48x^3-56x+25x^3-30x^2+35)+(-60x^4+48x^3+75x^3-60x^2)\)

\(\displaystyle (fg)'=(-40x^4+73x^3-30x^2-56x+35)+(-60x^4+123x^3-60x^2)\)

\(\displaystyle (fg)'=-100x^4+196x^3-90x^2-56x+35\)

Recall the product rule for differentiaion.

\(\displaystyle (fg)'=f'g+fg'\)

So we need to find the first derivative of each of our functions:

\(\displaystyle j(x)=5x^5- 7x\)

\(\displaystyle j'(x)=25x^4- 7\)

Recall that \(\displaystyle e^x\) is a strange one:

\(\displaystyle k(x)=e^x\)

\(\displaystyle k'(x)=e^x\)

Next, use the formula from up above.

\(\displaystyle (jk)'=(25x^4-7)(e^x)+(5x^5-7x)(e^x)\)

Expand and simplify.

\(\displaystyle (jk)'=25x^4e^x-7e^x+5x^5e^x-7xe^x\)

Rewrite in standard form and factor out an \(\displaystyle e^x\).

\(\displaystyle (jk)'=e^x(5x^5+25x^4-7x-7)\)

Step 1: Define \(\displaystyle f(x)\) and \(\displaystyle g(x)\):

\(\displaystyle f(x)=2x^2+x\)
\(\displaystyle g(x)=x^2-1\)
Step 2: Find \(\displaystyle f'(x)\) and \(\displaystyle g'(x)\).

\(\displaystyle f'(x)=4x+1\)
\(\displaystyle g'(x)=2x\)

Step 3: Define Product Rule:

Product Rule=\(\displaystyle f(x)g'(x)+f'(x)g(x)\).

Step 4: Substitute the functions for their places in the product rule formula

\(\displaystyle (2x^2+x)(2x)+(4x+1)(x^2-1)\)

Step 5: Expand:

\(\displaystyle 4x^3+2x^2+4x^3-4x+x^2-1\)

Step 6: Combine like terms:

Final answer: \(\displaystyle 8x^3+3x^2-4x-1\)

1) Starting Equation: \(\displaystyle y = 2x^3(2x^2)+2x+1{}\)

2) Simplifying:         \(\displaystyle y = 4x^5 + 2x + 1\)

3) Take the derivative, using the power rule.

                              \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = (5*4)x^{5-1} + (2*2x^{1-1}) + (0*1)\)

4) Simplify answer:  \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = 20x^4 + 2\)

Notes:

1) Easiest way to take the derivative is to simplify the equation first. In doing so, you should see that this is NOT an application for the chain rule. Although two variables are multiplied together, they are the same variable. The chain rule will give you the wrong answer.

2) Exponent Math... multiplied factors means you should add the exponents

3) Standard Power Rule. Bring the exponent down, multiplying it into the coefficient. Subtract 1 from the exponent. Constants go to 0.

4) Simplify the answer.

Step 1: We need to define the product rule. The product rule says \(\displaystyle f'g+g'f\). \(\displaystyle f'g\) is defined as the derivative of f(x) multiplied by g(x). In the question, the first parentheses is f(x) and the second parentheses is g(x).

Step 2: We will first calculate \(\displaystyle f'\) and \(\displaystyle g'\). To find the derivative of any term, we do one the following rules:

1) All terms with exponents (positive and negative) of the original equation are dropped down and multiplied by the coefficient of that term that you are working on. The exponent that gets written after taking the derivative is 1 less than (can also be thought of as (x-1, where x is the exponent that was dropped).
2) The derivative of a term ax, \(\displaystyle a\ne 0\), is just the value \(\displaystyle a\), which is the coefficient of the term.
3) The derivative of any constant term, that is any term that does not have a variable next to it, is always \(\displaystyle 0\).

Step 3: We will take derivative of \(\displaystyle f(x)\) first.

\(\displaystyle f(x)=2x^2+3x-4\). Let us take the derivative of each and every term and then add everything back together. We denote (') as derivative.

\(\displaystyle (2x^2)'=((2*2)x^{2-1})=4x\). We are using rule 1 here (listed above). The two from the exponent dropped down and was multiplied by the coefficient of that term. The exponent is 1 less than the exponent that was dropped down, which is why you see \(\displaystyle (2-1)\) in the exponent.
\(\displaystyle (3x)'=3\). We use rule 2 that was listed above. The derivative of this term is just the coefficient of that term, in this case, 3.
\(\displaystyle (-4)'=0\). We use rule 3. Since the derivative is \(\displaystyle 0\), we won't write it in the final equation for the derivative of \(\displaystyle f(x)\).

Let's put everything together:
\(\displaystyle (f(x))'=4x+3\)

Step 4: We will take derivative of g(x).

\(\displaystyle (6x^3)'=(6\cdot 3)x^{3-1}=18x^{2}\)
\(\displaystyle (-12x^2)'=(-12\cdot 2)x^{2-1}=-24x\)
\(\displaystyle (4)'=0\)

So, \(\displaystyle (g(x))'=18x^2-24x\).

Step 5: Now that we have found the derivatives, let's substitute all the equations into the formula for product rule.

\(\displaystyle fg'+f'g=[(2x^2+3x-4)(18x^2-24x)]+[(4x+3)(6x^3-12x^2+4)]\)

Step 6: Let's find \(\displaystyle fg'\).

In the equation above, \(\displaystyle fg'=[(2x^2+3x-4)(18x^2-24x)]\). We will need to distribute and expand this multiplication.

When we expand, we get \(\displaystyle 36x^4-48x^3+54x^3-72x^{2}-72x^2+96x\). Let's simplify that expansion.

We will get: \(\displaystyle 36x^4+6x^3-144x^2+96x\).

Step 7: Let's find \(\displaystyle f'g\), which is defined as \(\displaystyle (4x+3)(6x^3-12x^2+4)\).

We will expand and simplify.

When we expand, we get: \(\displaystyle 24x^4-48x^3+16x+18x^3-36x^2+12\).

If we simplify, we get \(\displaystyle 24x^4-30x^3-36x^2+16x+12\).

Step 8: Add the two products together and simplify.

If we add and simplify, we get:

\(\displaystyle 60x^4-24x^3-180x^2+112x+12\). This is the final answer to the expansion of the product rule. 

Step 1: Define the two functions...

\(\displaystyle f(x)=x^2+1\)
\(\displaystyle g(x)=x+1\)

Step 2: Find the derivative of each function:

\(\displaystyle f'(x)=2x\)
\(\displaystyle g'(x)=1\)

Step 3: Define the Product Rule Formula... 

\(\displaystyle f'(x)g(x)+f(x)g'(x)\)

Step 4: Plug in the functions:

\(\displaystyle 2x(x+1)+(x^2+1)\cdot 1\)

Step 5: Expand and Simplify:

\(\displaystyle 2x(x+1)+(x^2+1)\cdot 1\)
\(\displaystyle 2x^2+2x+x^2+1\)
\(\displaystyle 3x^2+2x+1\)

The derivative of the product of \(\displaystyle f(x)\) and \(\displaystyle g(x)\) is \(\displaystyle 3x^2+2x+1\).

Write the quotient rule.

\(\displaystyle \frac{d}{dx}\left (\frac{f(x)}{g(x)} \right )=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\)

For the function \(\displaystyle y=\frac{ln(x)}{cos(x)}\), \(\displaystyle f(x)=ln(x)\) and \(\displaystyle g(x)=cos(x)\), \(\displaystyle f'(x)= \frac{1}{x}\) and \(\displaystyle g'(x)=-sin(x)\).

Substitute and solve for the derivative.

\(\displaystyle \frac{d}{dx}\left (\frac{f(x)}{g(x)} \right )=\frac{cos(x)(\frac{1}{x})-ln(x)(-sin(x))}{cos(x)^2}\)

\(\displaystyle \frac{d}{dx}\left (\frac{f(x)}{g(x)} \right )=\frac{\frac{cos(x)}{x}+ln(x)sin(x)}{cos(x)^2}\)

Reduce the first term.

\(\displaystyle y'=\frac{1}{xcos(x)}+\frac{ln(x)sin(x)}{cos(x)^2}\)

This question asks us to find the derivative of a quotient. Use the quotient rule:

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}\)

Start by finding \(\displaystyle f'\) and \(\displaystyle g'\).

\(\displaystyle f(x)=-13x^3+7x^2\)

\(\displaystyle f'(x)=-39x^2+14x\)

\(\displaystyle g(x)=x^9\)

\(\displaystyle g'(x)=9x^8\)

So we get:

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{(-39x^2+14x)(x^9)-(-13x^3+7x^2)(9x^8)}{(x^9)^2}\)

Whew, let's simplify

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{(-39x^{11}+14x^{10})-(-117x^{11}+63x^{10})}{x^{18}}\)

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{-39x^{11}+14x^{10}+117x^{11}-63x^{10}}{x^{18}}\)

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{78x^{11}-49x^{10}}{x^{18}}\)

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{78x-49}{x^{8}}\)

This question yields to application of the quotient rule:

 \(\displaystyle \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}\)

So find \(\displaystyle f'\) and \(\displaystyle g'\) to start:

\(\displaystyle f(x)=5x^3\)

\(\displaystyle f'(x)=15x^2\)

\(\displaystyle g(x)=e^x\)

\(\displaystyle g'(x)=e^x\)

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{15x^2\cdot e^x-5x^3\cdot e^x}{(e^x)^2}\)

\(\displaystyle \left(\frac{f}{g}\right)'=\frac{e^x(15x^2-5x^3)}{(e^x)(e^x)}=\frac{15x^2-5x^3}{e^x}\)

So our answer is:

\(\displaystyle \frac{15x^2-5x^3}{e^x}\)