We begin by rewriting our function in a more convenient way,

\(\displaystyle f(x) = \ln \left(\frac{6x^2}{\left(x+4 \right )} \right )= \ln \left(6x^2 \right )- \ln \left(x+4 \right )\).

This makes taking the derivative a little easier.  We can therefore write

\(\displaystyle f' (x) = \frac{2}{x}-\frac{1}{x+4}\), or \(\displaystyle f' (1) = 2 - \frac{1}{5} = \frac{9}{5}\).

By the chain rule, we can show

\(\displaystyle \frac{d}{dx} \sin \left(f(x) \right )= cos \left(f(x) \right ) \cdot f'(x)\), or in this case,

\(\displaystyle \frac{d}{dx} \sin \left( \pi x \right )= \cos \left( \pi x \right ) \cdot \pi\Rightarrow \cos \left(5 \pi \right ) \cdot \pi = -\pi\).

This derivative must be found using logarithmic differentiation.  Consider the following

\(\displaystyle \begin{align*} f(x)&=x^{x^2} \\ \log_x f&= x^2\\ \frac{\ln f}{\ln x}&=x^2 \\ \ln f &= x^2 \ln x \\ \frac{f'}{f}&=x+2x \ln x \\ f'&= \left(x + 2x \ln x \right )x^{x^2}= f'(x) = x^{x^2+1}\left(1+2 \ln x \right ) \end{align*}\)

This derivative is most easily done by logarithmic differentiation.  Consider the following

\(\displaystyle \begin{align*} f(x)&=7^{3x} \\ \ln f&=\ln 7 \cdot 3x \\ \frac{f'}{f}&=\ln 7 \cdot 3 \\ f'(x)&=7^{3x} \cdot 3 \ln 7 \\ f'(x)&=7^{3x} \cdot \ln 7^3 \end{align*}\)

Take the derivative of the cosine function.  This will also require chain rule, which is the derivative of the inner function.

\(\displaystyle \frac{dy}{dx} = -[-sin(5x+2)] \cdot 5\)

The answer is:  \(\displaystyle \frac{dy}{dx}= 5sin(5x+2)\)

This problem will require multiple chain rule.  Take the derivative of natural log, and then apply the inner function of the natural log, cosine, and then also apply the chain rule for \(\displaystyle 2x\).

\(\displaystyle \frac{d}{dx}ln(cos(2x)) = \frac{1}{cos(2x)} \cdot -sin(2x) \cdot 2\)

Combine this into one term and simplify.

\(\displaystyle -\frac{2sin(2x)}{cos(2x)} = -2tan(2x)\)

The answer is:  \(\displaystyle -2tan(2x)\)

The correct answer is \(\displaystyle (x+1)y'+y\).

Since we are taking the (full) derivative of an expression involving \(\displaystyle x\) and \(\displaystyle y\), we must use implicit differentiation. We proceed as follows

\(\displaystyle \frac{d}{dx}(yx+y)\) start.

\(\displaystyle =\frac{d}{dx}(yx)+\frac{d}{dx}(y)\)

\(\displaystyle =(xy'+(1)y)+y'\). Using \(\displaystyle \frac{d}{dx}y=y'\), and the Product Rule.

\(\displaystyle =(1+x)y' +y\). Factor

The definition of the derivative is

\(\displaystyle \lim_{h \to 0}\frac{f(a+h)-f(a)}{h} = f'(a)\).

Equating

\(\displaystyle \lim_{h \to 0}\frac{12(\frac{1}{2}+h)^{12}+12(\frac{1}{2})^{12}}{h}\), to the definition of the derivative, we have

\(\displaystyle \lim_{h \to 0}\frac{12(\frac{1}{2}+h)^{12}+12(\frac{1}{2})^{12}}{h} = \frac{d}{dx}(12x^{12})|_{x=1/2}=144x^{11}|_{x=1/2}\)

\(\displaystyle =144(\frac{1}{2})^{11}=\frac{144}{2048} =\frac{9}{128}\).

Step 1: Take the derivative of the first term: \(\displaystyle 5x^4\)

\(\displaystyle (5x^4)'=20x^{4-1}=20x^3\) via the power rule.
Step 2: Take the derivative of the next term: \(\displaystyle 2x^2\)
\(\displaystyle (2x^2)'=4x^{2-1}=4x\), again using the product rule.
Step 3: Take the derivative of the last term \(\displaystyle \frac {3}{x}\).

We will have to use Quotient Rule here.

Find the derivative of f(x) and g(x):

\(\displaystyle f'(x)=0\), \(\displaystyle g'(x)=1\).

Use the formula: \(\displaystyle \frac {f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\)

\(\displaystyle \frac {0(x)-1(3)}{(x)^2}=\frac {-3}{x^2}\).

Step 4: Put all the final terms from highest to lowest degree...

We get:\(\displaystyle 20x^3+4x-\frac {3}{x^2}\).