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 $\displaystyle f(x)=xe^{4x}$ on the interval $\displaystyle (0,1)$

$\displaystyle f(0)=(0)e^{4(0)}=0$

$\displaystyle f(1)=(1)e^{4(1)}=54.6$

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

$\displaystyle \frac{54.6-0}{1-0}=54.6$

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

Derivative of an exponential: 

$\displaystyle d[e^u]=e^udu$

Derivative of a natural log: 

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

Product rule: $\displaystyle d[uv]=udv+vdu$

$\displaystyle f'(x)=e^{4x}+4xe^{4x}$

$\displaystyle e^{4x}+4xe^{4x}=54.6$

Using a calculator, we find the solution $\displaystyle x=0.67$, which fits within the interval $\displaystyle (0,1)$, satisfying the mean value theorem.

Assuming f(x) is continuous and differentiable for all values of x, what can be said about its graph at the point $\displaystyle x=5$ if we know that $\displaystyle f'(5)>0$ ?

We are told about a first derivative and asked to consider the original function. Recall that anywhere the first derivative is negative, the original function is decreasing. Anywhere the first derivative is positive, the original function is increasing.

We are told that $\displaystyle f'(5)>0$. In other words, the first derivative is positive.

This means our original function must be increasing.

f(x) is increasing when $\displaystyle x=5$

 $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, and the derivative of a constant is 0, so

$\displaystyle f (x) = 7x^{2} + 6x - 9$

$\displaystyle f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$\displaystyle f ' (x) = 14\cdot x^{1} + 6$

$\displaystyle f ' (x) = 14x + 6$

First, find the derivative $\displaystyle f'(x)$ of $\displaystyle f(x)= 8x^{3}-7x^{2}+11$.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, and the derivative of a constant is 0, so

$\displaystyle f(x)= 8x^{3}-7x^{2}+11$

$\displaystyle f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$\displaystyle f'(x)= 24x^{2}-14x^{1}$

$\displaystyle f'(x)= 24x^{2}-14x$

Now, differentiate $\displaystyle f'(x)$ to get $\displaystyle f''(x)$.

$\displaystyle f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$\displaystyle f''(x)= 48x^{1}-14x^{0}$

$\displaystyle f''(x)= 48x-14$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$, so

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

Find the derivative of $\displaystyle x^{777}$, then find the derivative of that expression.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, so

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (x^{777} \right )= 777\cdot x^{777-1} = 777x^{776}$

$\displaystyle \frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 777x^{776} \right )= 777\cdot776\cdot x^{776-1} = 602,952x^{775}$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, and the derivative of a constant is 0, so

$\displaystyle f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

$\displaystyle f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0$

$\displaystyle f'(x)= 2.4x^{2}-1x^{1}$

$\displaystyle f'(x) = 2.4x^{2}-x$

First, find the derivative $\displaystyle f'(x)$ of $\displaystyle f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$.

Recall that $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, and the derivative of a constant is 0.

$\displaystyle f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$\displaystyle f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate $\displaystyle f'(x)$ to get $\displaystyle f''(x)$.

$\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$\displaystyle f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$\displaystyle f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$\displaystyle f''(x) =10.8x^{2} + 4.2x$

We can use the (first part of) the Fundemental Theorem of Calculus to "cancel out" the integral.

$\displaystyle y = \int_{0}^{x^2} e^{-t^2}dt$. Start

$\displaystyle y' = \frac{d}{dx}\int_{0}^{x^2} e^{-t^2}dt$. Take the derivative of both sides with respect to $\displaystyle x$.

To "cancel out" the integral and the derivative sign, verify that the lower bound on the integral is a constant (It's $\displaystyle 0$ in this case), and that the upper limit of the integral is a function of $\displaystyle x$, (it's $\displaystyle x^2$ in this case).

Afterward, plug $\displaystyle x^2$ in for $\displaystyle t$, and ultilize the Chain Rule to complete using the Fundemental Theorem of Calculus.

$\displaystyle y' = e^{-(x^2)^2}(x^2)' = 2xe^{-x^4}$. 

The derivative of inverse cosine is:

$\displaystyle \frac{d}{dx} cos^-^1(x) = -\frac{1}{\sqrt{1-x^2}}$

The derivative of cosine is:

$\displaystyle \frac{d}{dx} cos(x)=-sin(x)$

Combine the two terms into one term.

$\displaystyle -\frac{1}{\sqrt{1-x^2}}-sin(x)$

$\displaystyle -\frac{1}{\sqrt{1-x^2}}-\frac{sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}= -\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$