Let \(\displaystyle g(x)=3x^2+2x\).

Then \(\displaystyle f(x) = g(x)^3\).

By the chain rule,

\(\displaystyle \frac{df}{dx}=\frac{df}{dg}\frac{dg}{dx}\)

\(\displaystyle \frac{df}{dg}=3g(x)^2\), \(\displaystyle \frac{dg}{dx}=6x+2\)

Plugging everything in we get

\(\displaystyle \frac{df}{dx}=3(3x^2+2x)^2(6x+2)=(18x+6)(3x^2+2x)^2\)

Let \(\displaystyle g(x)=x^2\) and \(\displaystyle h(x)=(e^x+ex^3)^{-1}\).

So \(\displaystyle f(x)=g(x)h(x)\).

By the product rule:

\(\displaystyle \frac{df}{dx}=g'h+h'g\)

Where \(\displaystyle h'=\frac{dh}{dx}\) and \(\displaystyle g'=\frac{dg}{dx}\).

Therefore,

\(\displaystyle g'=2x\)

\(\displaystyle h'=-(e^x+3x^3)^{-2}(e^x+9x^2)\)

Plugging everything in and simplifying we get:

\(\displaystyle \frac{df}{dx}=\frac{2x(e^x+3x^3)-x^2(e^x+9x^2)}{(e^x+3x^3)^2}\)

We can simplify the function by using the properties of logarithms.

\(\displaystyle f(x)=ln[x^3]+ln[e^x]=3ln[x]+x\)

With the simplified form, we can now find the derivative using the power rule which states,

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\). 

Also we will need to use the product rule which is,

\(\displaystyle f(x)=g(x)h(x)\rightarrow f'(x)=g(x)h'(x)+h(x)g'(x)\).

Remember that the derivative of \(\displaystyle ln(x)\rightarrow \frac{1}{x}\).

Applying these rules we find the derivative to be as follows.

\(\displaystyle \frac{df}{dx}=\frac{d}{dx}[3ln[x]+x]=\frac{3}{x}+1\)

For a function of the form \(\displaystyle f(x)=a^x\) the derivative is by definition:

\(\displaystyle \frac{d}{dx}f(x)=f'(x)=ln(a)a^x\).

Therefore,

\(\displaystyle f(x)=3^x\rightarrow f'(x)=3^xln(x)\).

Recall that, 

\(\displaystyle sin^2(x)=sin(x)sin(x)\)

Using the product rule

\(\displaystyle \frac{d}{dx}sin^2(x)=cos(x)sin(x)+sin(x)cos(x)=2cos(x)sin(x)\)

To compute the differential of the function we will need to use the power rule which states,

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

Applying the power rule we get: 

\(\displaystyle \frac{dy}{dt}=18t+24\)

From here solve for dy: 

\(\displaystyle dy=(18t+24)dt\)

Using the power rule,

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

the derivative of \(\displaystyle 7x^3\) becomes \(\displaystyle 21x^2\).

Using trigonometric identities, the derivative of \(\displaystyle cos(x)\) is \(\displaystyle -sin(x)\). 

Therefore, 

\(\displaystyle \frac{dy}{dx}=21x^2-sin(x)\)

\(\displaystyle dy=(21x^2-sin(x))dx\)

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

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