Take the derivative of each term. Remember that the derivative of y should be dy/dx. 

\(\displaystyle \frac{d(y^2 + y +x =0)}{dx} = 2y\frac{dy}{dx}+\frac{dy}{dx}+1=0\)

Remember to follow through with the chain rule for \(\displaystyle y^2\). Have to multiply by the derivative of the inside of the squared, which is just \(\displaystyle \frac{dy}{dx}\).

Now isolate \(\displaystyle \frac{dy}{dx}\):

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

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

To make solving easier, take the natural log of both sides of the equations:

\(\displaystyle ln(y) = ln(x^x)\)

Use the natural log properties of \(\displaystyle ln(a^b) = bln(a)\)

\(\displaystyle ln(y) = xln(x)\)

Now take the derivative of both sides of the equation, note that the derivative of \(\displaystyle y\) is \(\displaystyle \frac{dy}{dx}\) in this case. Must also use the power rule for the \(\displaystyle xln(x)\) term. The general equation is \(\displaystyle \frac{d(ab)}{dx}\) = \(\displaystyle \frac{da}{dx}*b+a*\frac{db}{dx}\)

\(\displaystyle \frac{dy}{dx}\frac{1}{y}= (1)lnx+x(\frac{1}{x})\)

Now simplify:

\(\displaystyle \frac{dy}{dx}= y(lnx+1)\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is found by \(\displaystyle f'(x)= (C)x^{(C-1)}\).

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= 8x^{\frac{1}{8}} - 12x^{\frac{1}{2}} + 4x - 6\)

That is done by doing the following:

\(\displaystyle f'(x)= (\frac{1}{8})8x^{(\frac{1}{8}-1)} - (\frac{1}{2})12x^{(\frac{1}{2}-1)} + (1)4x^{(1-1)} - 0\)

Therefore, the answer is:

\(\displaystyle f'(x)= x^{-\frac{7}{8}} - 6x^{-\frac{1}{2}}+4\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is found by \(\displaystyle f'(x)= (C)x^{(C-1)}\).

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= 3x^{\frac{2}{3}} - x^{4} + 2\)

That is done by doing the following:

\(\displaystyle f'(x)= (\frac{2}{3})3x^{(\frac{2}{3}-1)} - (4)x^{(4-1)} + 0\)

Therefore, the answer is:

\(\displaystyle f'(x)= 2x^{-\frac{1}{3}} - 4x^3\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is \(\displaystyle f'(x)= (C)x^{(C-1)}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\)

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= 2e^x - 3x^\frac{1}{3} + 5x\)

That is done by doing the following:

\(\displaystyle f'(x)= (2)(e^x) - (\frac{1}{3})3x^{(\frac{1}{3}-1)} + (1)5x^{(1-1)}\)

Therefore, the answer is:

\(\displaystyle f'(x)= 2e^x - x^{-\frac{2}{3}}+5\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is \(\displaystyle f'(x)= (C)x^{(C-1)}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\)

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= 2ln(x) - 8x^3 +4x^6 -23\)

That is done by doing the following:

\(\displaystyle f'(x)= 2(\frac{1}{x}) - (3)8x^{(3-1)} +(6)4x^{(6-1)} -0\)

Therefore, the answer is:

\(\displaystyle f'(x)= \frac{2}{x} - 24x^2 + 24x^5\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is \(\displaystyle f'(x)= (C)x^{(C-1)}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\)

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= 3x^2 - 4ln(x) + 100x^{\frac{1}{100}}\)

That is done by doing the following:

\(\displaystyle f'(x)= (2)3x^{(2-1)} - (4)(\frac{1}{x}) + (\frac{1}{100})100x^{(\frac{1}{100}-1)}\)

Therefore, the answer is:

\(\displaystyle f'(x)= 6x - \frac{4}{x} + x^{-\frac{99}{100}}\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is \(\displaystyle f'(x)= (C)x^{(C-1)}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)=\sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= \sin x - \cos x\)

That is done by doing the following:

\(\displaystyle f'(x)= \cos x - (-\sin x)\)

Therefore, the answer is:

\(\displaystyle f'(x)=\cos x + \sin x\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is \(\displaystyle f'(x)= (C)x^{(C-1)}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)=\sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= \sec x - \cos x\)

That is done by doing the following:

\(\displaystyle f'(x)= (\sec x) (\tan x) - (-\sin x)\)

Therefore, the answer is:

\(\displaystyle f'(x)= \sec x \tan x +\sin x\)

In order to find the derivative of a given function, there are sets of rules you must follow:

If \(\displaystyle f(x)= C\)(constant), then the derivative is \(\displaystyle f'(x)= 0\).

If \(\displaystyle f(x)= x^{C}\), then the derivative is \(\displaystyle f'(x)= (C)x^{(C-1)}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)=\sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

In this case, we must find the derivative of the following:

\(\displaystyle f(x)= -\cot x + \tan x + 3x\)

That is done by doing the following:

\(\displaystyle f'(x)= -(-\csc^2 x) + \sec ^2 x + (1)3x^{(1-1)}\)

Therefore, the answer is:

\(\displaystyle f'(x)= \csc^2 x + \sec ^2 x + 3\)