We will use the fact that $\displaystyle e^{au} = ae^{au}$ to differentiate. Let $\displaystyle a=6$ and $\displaystyle u=x$. Substituing our values we can see the derivative will be $\displaystyle 6e^{6x}$.

Using the product rule, we determine the derivative of $\displaystyle f(x)\cdot g(x) = {f}'(x)\cdot g(x) + f(x)\cdot {g}'(x))$
Let $\displaystyle f(x) = x^{2}$ and $\displaystyle g(x) = \sin(x)$. We can see that $\displaystyle f{}'(x) = 2x$ and $\displaystyle g{}'(x) = \cos(x)$.

Plugging in our values into the product rule formula, we are left with the final derivative of $\displaystyle x (2 \sin(x)+x \cos(x))$.

According to the power rule, whenever we differentiate a constant value it will reduce to zero. Since the only term of our function is a constant, we can only differentiate  $\displaystyle 74 \rightarrow 0$.

Using the chain rule, we will differentiate the exponent of our exponential function, and then multiply our original function. Differentiating our exponent with the power rule will yield $\displaystyle 4x$. Using the chain rule we will multiply this by our original function resulting in $\displaystyle 4xe^{2x^{2}}$.

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $\displaystyle x^{2}$, will thus become $\displaystyle 2x$. The second term is a constant value, so according to the power rule this term will become $\displaystyle 0$.

Using the chain rule, we will determine the derivative of our function will be $\displaystyle \frac{d\log(u)}{du}\cdot \frac{du}{dx}$.

The derivative of the log function is $\displaystyle \frac{1}{x}$, and our second term of the chain rule will cancel out $\displaystyle \frac{1}{4x}\cdot 4=\frac{4}{4x}=\frac{1}{x}$.

Thus our derivative will be $\displaystyle \frac{1}{x}$.

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $\displaystyle x^{4}$, will thus become $\displaystyle 4x^{3}$. The second term $\displaystyle 3x^{3}$, will thus become $\displaystyle 9x^{2}$. The last term is $\displaystyle 4x$, will reduce to $\displaystyle 4$.

According to the quotient rule, the derivative of ,

$\displaystyle \frac{f(x)}{g(x)} = \frac{{f}'(x)g(x) - f(x){g}'(x))}{(g(x))^{2}}$.

We will let $\displaystyle f(x) = e^{x} , f{}'(x) = e^{x} , g(x) = x^{2}$ and $\displaystyle g{}'(x) = 2x$
Plugging all of our values into the quotient rule formula we come to a final solution of :

$\displaystyle \frac{e^x (x-2)}{x^3}$. 

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $\displaystyle x^{3}$, will thus become $\displaystyle 3x^{2}$. The second term $\displaystyle 3x^{2}$, will thus become $\displaystyle 6x$.

Rewrite $\displaystyle \frac{dy}{dx}= \frac{y}{2}$ by multiply the $\displaystyle dx$ on both sides, and dividing $\displaystyle y$ on both sides of the equation.

$\displaystyle \frac{1}{y}dy= \frac{1}{2}dx$

Integrate both sides of the equation and solve for y.

$\displaystyle \int\frac{1}{y}dy= \int\frac{1}{2}dx$

$\displaystyle ln\left | y\right |=\frac{1}{2}x+C$

$\displaystyle \left | y\right |= e^0^.^5^xe^C$

$\displaystyle y=Ce^0^.^5^x$