By Fundamental Theorem of Calculus,

 $\displaystyle \frac{dH}{dx}=\frac{\mathrm{d} }{\mathrm{d} x} \int_{2}^{x} \ln (\sqrt{t^2+16}) \ dt = H'(x) \cdot \frac{d}{dx}x = \ln(\sqrt{x^2+16}) \cdot 1 =\ln(\sqrt{x^2+16}).$

Since $\displaystyle \frac{5}{27}e^2$ is a constant, its derivative is $\displaystyle 0$. Therefore, $\displaystyle f'(x) = \cos x + 2x.$

To begin with write the information of the problem in mathematical terms. We're told the sum of two numbers is fifty:

$\displaystyle x+y=50$

Their product is then:

$\displaystyle f(x,y)=xy$

We're uncertain currently about what these two numbers are, but we can relate them:

$\displaystyle y=50-x$

So that we can simplify the product function:

$\displaystyle f(x)=x(50-x)$

$\displaystyle f(x)=50x-x^2$

Maxima and minima occur where the derivative of a function is zero. Take the derivative of this function:

$\displaystyle f'(x)=50-2x$

And set it to zero:

$\displaystyle 50-2x=0$

$\displaystyle -2x=-50$

$\displaystyle x=25$

$\displaystyle y=50-x=50-25=25$

You can validate that this is a minimum due to the derivative function having positive values before this point, and negative values after this point.

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

$\displaystyle f(x)=(x^2+1)^5$ on the interval $\displaystyle (0,2)$

$\displaystyle f(0)=(0^2+1)^5=1$

$\displaystyle f(2)=(2^2+1)^5=3125$

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

$\displaystyle \frac{3125-1}{2-0}=1562$

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

$\displaystyle f'(x)=5(2x)(x^2+1)^4$

$\displaystyle 5(2x)(x^2+1)^4=1562$

$\displaystyle x=1.484$

This value valls within the interval $\displaystyle (0,2)$, validating the mean value theorem.

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

$\displaystyle f(x)=15x^3+23x^2-8x$ on the interval $\displaystyle (1,1.01)$

$\displaystyle f(1)=15(1)^3+23(1)^2-8(1)=30$

$\displaystyle f(1.01)=15(1.01)^3+23(1.01)^2-8(1.01)=30.836815$

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

$\displaystyle \frac{30.836815-30}{1.01-1}=83.6815$

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

$\displaystyle f'(x)=45x^2+46x-8$

$\displaystyle 45x^2+46x-8=83.6815$

$\displaystyle x=-2.02722,1.005$

To validate the mean value theorem, a solution must fall within the interval used, and $\displaystyle 1.005$ does in fact fall within $\displaystyle (1,1.01)$

Use the power rule to find the derivative.

$\displaystyle \frac{d}{dx}2x^2=4x$

$\displaystyle \frac{d}{dx}4x=4$

$\displaystyle \frac{d}{dx}-6=0$

Thus, the derivative is $\displaystyle 4x+4$

Use the power rule to find the derivative. 

$\displaystyle \frac{d}{dx}6x^3=18x^2$

First, find the derivative using the power rule:

$\displaystyle \frac{d}{dx}6x^3=18x^2$

Now, substitute 3 for x.

$\displaystyle 18(3^2)=162$

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

$\displaystyle f(x)=9^{\cos(x^2)}$ on the interval $\displaystyle (0,0.3)$

$\displaystyle f(0)=9^{\cos(0^2)}=9$

$\displaystyle f(0.3)=9^{\cos(0.3^2)}=8.9203$

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

$\displaystyle \frac{8.9203-9}{0.3-0}=-0.266$

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

Derivative of an exponential: 

$\displaystyle d[a^u]=a^udu\ln(a)$

Trigonometric derivative: 

$\displaystyle d[\cos(u)]=-\sin(u)du$

$\displaystyle f'(x)=-2x\sin(x^2)9^{\cos(x^2)}\ln(9)$

$\displaystyle -2x\sin(x^2)9^{\cos(x^2)}\ln(9)=-0.266$

Using a solver, there are multiple solutions which satisfy this equation:

$\displaystyle x=-3.54,-3.10,-2.51,-1.85,0.19,1.69,2.51,3.04$

However, to satisfy the mean value theorem, a solution must fall within the specified interval. $\displaystyle 0.19$ falls within $\displaystyle (0,0.3)$.

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.

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

Using our interval and function definitions, along with the derivative value of $\displaystyle f'(c)$ given, we'll in turn wish to solve for $\displaystyle b$ to validate the mean value theorem:

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$

$\displaystyle -1=\frac{(10b-b^2)-(10(2)-2^2)}{b-2}$

$\displaystyle 2-b=10b-b^2-16$

$\displaystyle b^2-11b+18=0$

$\displaystyle b=9$

(Note that the solution $\displaystyle b=2$ would not work because it would eliminate the interval as $\displaystyle a=2$, and it would lead to a zero denominator in the working equation.)