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.

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

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

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

\(\displaystyle y=Ce^0^.^5^x\)