When differentiating a function, you must understand how to find the derivative of different types of functions:

If \(\displaystyle f(x) = c\) (constant), then \(\displaystyle f'(x) = 0\).

If \(\displaystyle \large f(x) = e^{}x\), then \(\displaystyle \large f'(x) = e^{x}\)^.

If\(\displaystyle f(x) = x^{n}\), then \(\displaystyle f'(x) = (n)x^{n-1}\).

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

There are multiple different rules that apply when differentiating different kinds of functions.

In this case, you find the derivative of each portion of the function:

\(\displaystyle f(x) = 2x^{5} + 3e^{x} - 45\)  \(\displaystyle \large \rightarrow\)  \(\displaystyle f'(x) = (5)2x^{(5-1)} + 3e^{x} - 0\)

And so the simplified answer is:

^{\(\displaystyle f'(x) = 10x^{4} + 3e^{x}\).}

When differentiating a function, you must understand how to find the derivative of different types of functions:

If \(\displaystyle f(x) = c\) (constant), then \(\displaystyle f'(x) = 0\).

If \(\displaystyle \large f(x) = e^{}x\), then \(\displaystyle \large f'(x) = e^{x}\)^.

If\(\displaystyle f(x) = x^{n}\), then \(\displaystyle f'(x) = (n)x^{n-1}\).

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

There are multiple different rules that apply when differentiating different kinds of functions.

In this case, you find the derivative of each portion of the function:

\(\displaystyle f(x) = 12\cos x - 3\sin x + 2x\)   \(\displaystyle \LARGE \rightarrow\)\(\displaystyle f'(x) = 12(-\sin x) - 3(\cos x) + (1)2x^{(1-1)}\)

And so the simplified answer is: \(\displaystyle f'(x) = -12\sin x - 3\cos x + 2\)

In order to find the second derivative, you must first find the first derivative. To find the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent. Therefore, the first derivative is: \(\displaystyle f{}'(x)=-8x^3+18x^2-2x+5\). Then, take the derivative of the first derivative to find your second derivative, which is: \(\displaystyle f{}''(x)=-24x^2+36x-2\).

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:

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