It would be difficult to differentiate this equation by isolating \(\displaystyle y\). Luckily, we don't have to.  Use \(\displaystyle \small \frac{dy}{dx}\) to represent the derivative of \(\displaystyle y\) with respect to \(\displaystyle x\) and follow the chain rule.

(Remember, \(\displaystyle \small \frac{dx}{dx}\) is the derivative of \(\displaystyle x\) with respect to \(\displaystyle x\), although it usually doesn't get written out because it is equal to 1. We'll write it out this time so you can see how implicit differentiation works.)

\(\displaystyle \small 8x^{2}+3y^{2}+6x+y-104 = 0\)

\(\displaystyle \small 16x\frac{dx}{dx}+6y\frac{dy}{dx}+6\frac{dx}{dx}+\frac{dy}{dx}=0\)

\(\displaystyle \small 16x(1)+6y\frac{dy}{dx}+6(1)+\frac{dy}{dx}=0\)

\(\displaystyle \small 16x+6y\frac{dy}{dx}+6+\frac{dy}{dx}=0\)

Now we need to isolate \(\displaystyle \small \frac{dy}{dx}\) by first putting all of these terms on the same side:

\(\displaystyle \small \small 6y\frac{dy}{dx}+\frac{dy}{dx}=-16x-6\)

\(\displaystyle \small \frac{dy}{dx}(6y+1)=-16x-6\)

\(\displaystyle \small \frac{dy}{dx}=\frac{-16x-6}{6y+1}\)

This is the equation for the derivative at any point on the curve. By substituting in (3, 2) from the original question, we can find the slope at that particular point:

\(\displaystyle \small \frac{dy}{dx}=\frac{-16(3)-6}{6(2)+1}=\frac{-48-6}{12+1}=\frac{-54}{13}\)

The derivative must be computed using the product rule.  Because the derivative of \(\displaystyle x^2\) brings a \(\displaystyle 2\) down as a coefficient, it can be combined with \(\displaystyle 2^x\) to give \(\displaystyle 2^{x+1}\)

The instantaneous rate of change of \(\displaystyle f\) at \(\displaystyle x = x _{0}\) is \(\displaystyle f ' (x _{0})\), so we will find \(\displaystyle f ' (x)\) and evaluate it at \(\displaystyle x = 3\).

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }\) for any positive \(\displaystyle a\), so 

\(\displaystyle f'(x) =\frac{\mathrm{d}}{\mathrm{d} x}\left [ \left ( \frac{1}{3} \right ) ^{x }\right ] = \ln \frac{1}{3} \cdot \left ( \frac{1}{3} \right) ^{x }= -\frac{ \ln 3}{3^{x}}\)

\(\displaystyle f'(3) =-\frac{ \ln 3}{3^{3}} = -\frac{ \ln 3}{27}\)

\(\displaystyle f(x) = 5 ^{x + 1} = 5 ^{1} \cdot 5 ^{x } = 5 \cdot 5 ^{x }\)

Therefore, 

\(\displaystyle f'(x) = 5 \cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 5 ^{x } \right )\)

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }\) for any real \(\displaystyle a\), so \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left ( 5 ^{x } \right ) = \ln 5 \cdot 5 ^{x}\), and

\(\displaystyle f'(x) = 5 \ln 5 \cdot 5 ^{x} = \ln 5 \cdot 5 \cdot 5 ^{x} = \ln 5 \cdot 5 ^{x+1}\)

\(\displaystyle f(x) = 6 ^{x -2} = 6 ^{-2} \cdot 6 ^{x } = \frac{1}{36} \cdot 6 ^{x }\)

Therefore, 

\(\displaystyle f'(x) = \frac{1}{36} \cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 6 ^{x } \right )\)

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }\) for any positive \(\displaystyle a\), so \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left ( 6 ^{x } \right ) = \ln 6 \cdot 6 ^{x}\), and

\(\displaystyle f'(x) = \frac{1}{36} \ln 6 \cdot 6 ^{x} = \ln 6 \cdot \frac{1}{36} \cdot 6 ^{x}\)

\(\displaystyle f'(x) = \ln 6 \cdot 6 ^{x-2}\)

\(\displaystyle f'(4) = \ln 6 \cdot 6 ^{4-2} = \ln 6 \cdot 6 ^{2} = 36 \ln 6\)

The derivative of \(\displaystyle \sin (x)\) is\(\displaystyle \cos (x)\). It is probably best to memorize this fact (the proof follows from the difference quotient definition of a derivative).

Our function 

\(\displaystyle f(x) = 3 \;\textup{sin}(x)\)

the factor of 3 does not change when we differentiate, therefore the answer is

\(\displaystyle f(x) = 3 \;\textup{cos}(x)\)

The derivative of a sine function does not follow the power rule. It is one that should be memorized.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)\).

The derivatives of trig functions must be memorized. The first derivative is:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)\).

To find the second derivative, we take the derivative of our result.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)\).

Therefore, the second derivative will be \(\displaystyle -\sin(x)\).

Use the Chain Rule.

Set \(\displaystyle u = x^{2}\) and substitute.

\(\displaystyle f(x) = \tan u\)

\(\displaystyle \frac{\mathrm{d} f }{\mathrm{d} u} = \frac{\mathrm{d} }{\mathrm{d} u} \tan u = \sec^{2} u\)

\(\displaystyle \frac{\mathrm{d} u }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} x^{2} = 2x^{2-1} =2x\)

\(\displaystyle \frac{\mathrm{d} f }{\mathrm{d} x} = \frac{\mathrm{d} f }{\mathrm{d} u} \cdot \frac{\mathrm{d} u }{\mathrm{d} x}= \sec^{2} u \cdot 2x = 2x \sec^{2} x^{2}\)

Since this function is a polynomial, we take the derivative of each term separately.

From the power rule, the derivative of 

\(\displaystyle x^3\)

is simply

\(\displaystyle 3x^2\)

We can rewrite \(\displaystyle \frac{1}{x}\) as

\(\displaystyle x^{-1}\)

and using the power rule again, we get a derivative of

\(\displaystyle -x^{-2}\) or \(\displaystyle -\frac{1}{x^2}\)

So the answer is

\(\displaystyle f'(x) = 3x^2-\frac{1}{x^2}\)