The differential of $\displaystyle y$ is $\displaystyle dy$.

To find the differential of the right side of the equation, take the derivative of each term as you apply the product rule.

The product rule is:

$\displaystyle (second\quad term)\cdot d(first\quad term)+(first\quad term)\cdot d(second\quad term)$, so applying that rule to the equation yields:

 $\displaystyle dy=(2xcos(x)+x^2sin(x))dx$

The differential of $\displaystyle y$ is $\displaystyle dy$.

To find the differential of the right side of the equation, take the derivative of each term as you apply the product rule.

The product rule is: 

$\displaystyle (second\quad term)\cdot d(first\quad term) + (first\quad term)\cdot d(second\quad term)$, so applying that rule to the equation yields: 

$\displaystyle dy=(3x^{2}e^{x}+x^{3}e^{x}})dx$

The differential of $\displaystyle y$ is $\displaystyle dy$.

To find the differential of the right side of the equation, take the derivative of each term as follows.

The derivative of anything in the form of $\displaystyle x^n$ is $\displaystyle n{x^{n-1}}$, and the derivative of $\displaystyle -e^{-x}$is $\displaystyle e^{-x}$ so applying that rule to all of the terms yields: 

$\displaystyle dy=(12x^{2}+e^{-x})dx$

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, 

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

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