The mean value theorem states that for a planar arc passing through a starting and endpoint \(\displaystyle (a,b);a< b\), there exists at a minimum one point, \(\displaystyle c\), within the interval \(\displaystyle (a,b)\) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

\(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b\)

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of \(\displaystyle f(x)=ln(tan(x))\) on the interval \(\displaystyle (\frac{\pi}{4},\frac{\pi}{3})\)

\(\displaystyle f(\frac{\pi}{4})=ln(tan(\frac{\pi}{4}))=0\)

\(\displaystyle f(\frac{\pi}{3})=ln(tan(\frac{\pi}{3}))=0.549\)

Then take the difference of the two and divide by the interval.

\(\displaystyle \frac{0.549-0}{\frac{\pi}{3}-\frac{\pi}{4}}=2.097\)

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of a natural log: 

\(\displaystyle d[ln(u)]=\frac{du}{u}\)

Trigonometric derivative: 

\(\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}\)

\(\displaystyle f'(x)=\frac{1}{cos^2(x)tan(x)}=\frac{1}{cos(x)sin(x)}\)

\(\displaystyle \frac{1}{cos(x)sin(x)}=2.097\)

Using a calculator, we find the solution \(\displaystyle x=0.938\), which fits within the interval \(\displaystyle (\frac{\pi}{4},\frac{\pi}{3})\), satisfying the mean value theorem.

Use the power rule to find the derivative. 

\(\displaystyle \frac{d}{dx}x^3=3x^2\)

\(\displaystyle \frac{d}{dx}-3x^2=-6x\)

\(\displaystyle \frac{d}{dx}2x=2\)

Thus, the answer is \(\displaystyle 3x^2-6x+2\)

Use the power rule to find the derivative. 

\(\displaystyle \frac{d}{dx}x^3=3x^2\)

\(\displaystyle \frac{d}{dx}-3x^2=-6x\)

\(\displaystyle \frac{d}{dx}2x=2\)

Thus, the answer is \(\displaystyle 3x^2-6x+2\)

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

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

If \(\displaystyle f(x)= e^x\), then 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. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x)) g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= 2x^2 + \tan x - (2x-4)^2\)

That is done by doing the following: \(\displaystyle f'(x)= (2)2x^{2-1} + \sec^2x - (2)(2x-4)^{2-1} ( (1) 2x^{1-1} - 0 )\)

Therefore, the answer is: \(\displaystyle f'(x)= \sec^2x - 4x+16\)

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

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

If \(\displaystyle f(x)= e^x\), then 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. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x)) g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= \cos(\frac{1}{4}x^4 +3x)\)

That is done by doing the following: \(\displaystyle f'(x)= -\sin(\frac{1}{4}x^4 +3x) ((4)\frac{1}{4}x^{4-1} +(1)3x^{1-1})\)

Therefore, the answer is: \(\displaystyle f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)\)

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

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

If \(\displaystyle f(x)= e^x\), then 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. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x)) g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= (ln(2x))^2\)

That is done by doing the following: \(\displaystyle f'(x)= (2)(ln(2x))^{2-1} (\frac{1}{2x})(2)\)

Therefore, the answer is: \(\displaystyle f'(x)= ln(2x) (\frac{2}{x})\)

Instead of using FOIL to get a polynomial, we can use a special derivative rule, where we multiply the derivative of expression 1 by expression 2 and then add it to the product of teh derivative of expression 2 by expression 1: \(\displaystyle 2(3x^2+2)+6x(2x-1)\). Simplify to get your answer of: \(\displaystyle f{}'(x)=18x^2-6x+4\).

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

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

If \(\displaystyle f(x)= e^x\), then 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. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x)) g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= 2(x+2)^2 + 3x\)

That is done by doing the following: \(\displaystyle f'(x)= 2(2)(x+2)^{2-1}((1)x^{1-1}+0) + (1)3x^{1-1}\)

Therefore, the answer is: \(\displaystyle f'(x)= 4x+11\)

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

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

If \(\displaystyle f(x)= e^x\), then 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. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x)) g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= 2(x+2)^2 + 3x\)

That is done by doing the following: \(\displaystyle f'(x)= 2(2)(x+2)^{2-1}((1)x^{1-1}+0) + (1)3x^{1-1}\)

Therefore, the answer is: \(\displaystyle f'(x)= 4x+11\)

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

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

If \(\displaystyle f(x)= e^x\), then 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. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x)) g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= 5 ln (5x^2-3x+1)\)

That is done by doing the following: \(\displaystyle f'(x)=( 5) \frac{1}{5x^2+3x-1} ((2)5x^{2-1} + (1)3x^{1-1} - 0)\)

Therefore, the answer is: \(\displaystyle f'(x)=\frac{50x+ 15}{5x^2+3x-1}\)