To solve this problem, you must use the product rule of finding derivatives.

For any function \(\displaystyle y=ab\), \(\displaystyle y'=a(b')+b(a')\).

In this problem, the product rule yields 

\(\displaystyle a=sin(x), a'=cos(x)\)

\(\displaystyle b=cos(x), b'=-sin(x)\)

\(\displaystyle \frac{dy}{dx}=sin(x)\cdot(-sin(x))+cos(x)\cdot cos(x)\)

\(\displaystyle dy=[-sin^2(x)+cos^2(x)]dx\).

This differential can be found by utilizing the power rule,

\(\displaystyle y=x^n \rightarrow dy=nx^{n-1}dx\).

The original equation is \(\displaystyle y=2+3x-x^2\).

Using the power rule on each term we see that the derivative of \(\displaystyle 2\) is \(\displaystyle 0\). The derivative of a constant is always zero.

The dervative of \(\displaystyle 3x\) is

\(\displaystyle 1\cdot3 x^{1-1}=3x^0=3\).

The derivative of \(\displaystyle -x^2\) is

\(\displaystyle -1\cdot 2x^{2-1}=-2x^1=-2x\).

Thus, 

\(\displaystyle \frac{dy}{dx}=3-2x\)

Multiply \(\displaystyle dx\) to the right side to get the final solution. 

\(\displaystyle dy=(3-2x)dx\)

Using the power rule, we can find the derivative of each part of the function. The power rule states to multiply the coefficient of the term by the exponent then decrease the exponent by one.

The derivative of \(\displaystyle 4x^3\) is \(\displaystyle 12x^2\).

The derivative of \(\displaystyle 8x^2\) is \(\displaystyle 16x\).

The derivative of \(\displaystyle 32x\) is \(\displaystyle 32\).

And finally, because \(\displaystyle \pi\) is a constant, the derivative of \(\displaystyle \pi\) is \(\displaystyle 0\).

Thus, when we add the parts together, the derivative is 

\(\displaystyle \frac{dy}{dx}=12x^2+16x+32\)

and

\(\displaystyle dy=(12x^2+16x+32)dx\)

To solve this problem, you may use the quotient rule for finding derivatives. The quotient rule stipulates that for a function 

\(\displaystyle y=\frac{a}{b}\),  \(\displaystyle y'=\frac{ba'-ab'}{b^2}\).

In this problem, \(\displaystyle a=x^2\) and \(\displaystyle b=2\).

Thus, 

\(\displaystyle \frac{dy}{dx}=\frac{2(2x)-x^2(0)}{4}\) 

\(\displaystyle \frac{dy}{dx}=\frac{4x}{4}\)  

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

Therefore, 

\(\displaystyle \frac{dy}{dx}=x\) and \(\displaystyle dy=xdx\).

Using the power rule, we can find the derivative of each part of the function. When using the power rule you multiply the coefficient by the exponent then decrease the exponent by one.

The derivative of \(\displaystyle \frac{1}{2}g^2\) is \(\displaystyle \frac{1}{2}\cdot 2g^{2-1}=g\).

The derivative of \(\displaystyle \frac{3}{5}g^3\) is \(\displaystyle \frac{9}{5}g^2\).

The derivative of \(\displaystyle \frac{2}{3}g\) is \(\displaystyle \frac{2}{3}\). 

When these derivatives are added together, 

\(\displaystyle \frac{dy}{dg}=g+\frac{9}{5}g^2+\frac{2}{3}\).

Thus, 

\(\displaystyle dy=\left(g+\frac{9}{5}g^2+\frac{2}{3}\right)dg\)

Use the quotient rule to find this answer. The quotient rule dictates that for a function 

\(\displaystyle y=\frac{a}{b}\), \(\displaystyle y'=\frac{b(a')-a(b')}{b^2}\).

For this particular question, \(\displaystyle a=2x\) and \(\displaystyle b=x-1\).

Apply the quotient rule to this function:

 \(\displaystyle \frac{dy}{dx}=\frac{2x-2-2x}{(x-1)^2}=-\frac{2}{(x-1)^2}\)

\(\displaystyle dy=\left(-\frac{2}{(x-1)^2}\right)dx\)

For finding the derivative of a root, it is helpful to turn the root into a power.

For example, in this problem it is helpful to turn the \(\displaystyle \sqrt{x}\) into \(\displaystyle x^\frac{1}{2}\).

Now, we can easily apply the power rule,

\(\displaystyle y=x^n\rightarrow dy=nx^{n-1}dx\)

which yields the answer 

\(\displaystyle dy=\left(\frac{1}{2\sqrt{x}}\right)dx\).

Using the power rule, we can solve this problem. The power rule states to multiply the coefficient with the exponent of the term then decrease the exponent by one.

The derivative of \(\displaystyle 12x^2\) is \(\displaystyle 24x\).

The derivative of \(\displaystyle \sqrt{x}\) is \(\displaystyle \frac{1}{2\sqrt{x}}\).

Thus, 

\(\displaystyle \frac{dy}{dx}=24x-\frac{1}{2\sqrt{x}}\) and 

\(\displaystyle dy=\left(24x-\frac{1}{2\sqrt{x}}\right)dx\)

Use the power rule to differentiate this function. The power rule states to multiply the coefficient by the exponent then decrease the exponent by one.

The derivative of \(\displaystyle x^2\) is \(\displaystyle 2x\).

The derivative of \(\displaystyle 3x\) is \(\displaystyle 3\).

The derivative of \(\displaystyle 4x^3\) is \(\displaystyle 12x^2\).

Thus, 

\(\displaystyle \frac{dy}{dx}=2x+3+12x^2\) and \(\displaystyle dy=(2x+3+12x^2)dx\)

Use the quotient rule to find the solution to this problem. The quotient rule stipulates that for a function 

\(\displaystyle y=\frac{a}{b}\), \(\displaystyle y'=\frac{b(a')-a(b')}{b^2}\)

In this problem, \(\displaystyle a=6x^2\) and \(\displaystyle b=2-x\).

Apply the quotient rule: 

\(\displaystyle \frac{dy}{dx}=\frac{2-x(12x)-6x^2(-1)}{(2-x)^2}\)

\(\displaystyle \frac{dy}{dx}=\frac{24x-6x^2}{(2-x)^2}\)

\(\displaystyle dy=\left(\frac{24x-6x^2}{(2-x)^2}\right)dx\)