In this case we have the sum and difference of four terms and so we will differentiate each of the terms using the first property from above and then put them back together with the proper sign. Also, for each term with a multiplicative constant remember that all we need to do is “factor” the constant out (using the second property) and then do the derivative.

\[\begin{align*}f'\left( x \right) & = 15\left( {100} \right){x^{99}} - 3\left( {12} \right){x^{11}} + 5\left( 1 \right){x^0} - 0\\ & = 1500{x^{99}} - 36{x^{11}} + 5\end{align*}\]

Notice that in the third term the exponent was a one and so upon subtracting 1 from the original exponent we get a new exponent of zero. Now recall that \({x^0} = 1\). Don’t forget to do any basic arithmetic that needs to be done such as any multiplication and/or division in the coefficients.

The point of this problem is to make sure that you deal with negative exponents correctly. Here is the derivative.

\[\begin{align*}g'\left( t \right) & = 2\left( 6 \right){t^5} + 7\left( { - 6} \right){t^{ - 7}}\\ & = 12{t^5} - 42{t^{ - 7}}\end{align*}\]

Make sure that you correctly deal with the exponents in these cases, especially the negative exponents. It is an easy mistake to “go the other way” when subtracting one off from a negative exponent and get \( - 6{t^{ - 5}}\) instead of the correct \( - 6{t^{ - 7}}\).

Now in this function the second term is not correctly set up for us to use the power rule. The power rule requires that the term be a variable to a power only and the term must be in the numerator. So, prior to differentiating we first need to rewrite the second term into a form that we can deal with.

\[y = 8{z^3} - \frac{1}{3}{z^{ - 5}} + z - 23\]

Note that we left the 3 in the denominator and only moved the variable up to the numerator. Remember that the only thing that gets an exponent is the term that is immediately to the left of the exponent. If we’d wanted the three to come up as well we’d have written,

\[\frac{1}{{{{\left( {3z} \right)}^5}}}\]

so be careful with this! It’s a very common mistake to bring the 3 up into the numerator as well at this stage.

Now that we’ve gotten the function rewritten into a proper form that allows us to use the Power Rule we can differentiate the function. Here is the derivative for this part.

\[y' = 24{z^2} + \frac{5}{3}{z^{ - 6}} + 1\]

All of the terms in this function have roots in them. In order to use the power rule we need to first convert all the roots to fractional exponents. Again, remember that the Power Rule requires us to have a variable to a number and that it must be in the numerator of the term. Here is the function written in “proper” form.

\[\begin{align*}T\left( x \right) & = {x^{\frac{1}{2}}} + 9{\left( {{x^7}} \right)^{\frac{1}{3}}} - \frac{2}{{{{\left( {{x^2}} \right)}^{\frac{1}{5}}}}}\\ & = {x^{\frac{1}{2}}} + 9{x^{\frac{7}{3}}} - \frac{2}{{{x^{\frac{2}{5}}}}}\\ & = {x^{\frac{1}{2}}} + 9{x^{\frac{7}{3}}} - 2{x^{ - \frac{2}{5}}}\end{align*}\]

In the last two terms we combined the exponents. You should always do this with this kind of term. In a later section we will learn of a technique that would allow us to differentiate this term without combining exponents, however it will take significantly more work to do. Also, don’t forget to move the term in the denominator of the third term up to the numerator. We can now differentiate the function.

\[\begin{align*}T'\left( x \right) & = \frac{1}{2}{x^{ - \,\,\frac{1}{2}}} + 9\left( {\frac{7}{3}} \right){x^{\frac{4}{3}}} - 2\left( { - \frac{2}{5}} \right){x^{ - \,\,\frac{7}{5}}}\\ & = \frac{1}{2}{x^{ - \,\,\frac{1}{2}}} + \frac{{63}}{3}{x^{\frac{4}{3}}} + \frac{4}{5}{x^{ - \,\,\frac{7}{5}}}\end{align*}\]

Make sure that you can deal with fractional exponents. You will see a lot of them in this class.

In all of the previous examples the exponents have been nice integers or fractions. That is usually what we’ll see in this class. However, the exponent only needs to be a number so don’t get excited about problems like this one. They work exactly the same.

\[h'\left( x \right) = \pi {x^{\pi - 1}} - \sqrt 2 {x^{\sqrt 2 - 1}}\]

The answer is a little messy and we won’t reduce the exponents down to decimals. However, this problem is not terribly difficult it just looks that way initially.

In this function we can’t just differentiate the first term, differentiate the second term and then multiply the two back together. That just won’t work. We will discuss this in detail in the next section so if you’re not sure you believe that hold on for a bit and we’ll be looking at that soon as well as showing you an example of why it won’t work.

It is still possible to do this derivative however. All that we need to do is convert the radical to fractional exponents (as we should anyway) and then multiply this through the parenthesis.

\[y = {x^{\frac{2}{3}}}\left( {2x - {x^2}} \right) = 2{x^{\frac{5}{3}}} - {x^{\frac{8}{3}}}\]

Now we can differentiate the function.

\[y' = \frac{{10}}{3}{x^{\frac{2}{3}}} - \frac{8}{3}{x^{\frac{5}{3}}}\]

As with the first part we can’t just differentiate the numerator and the denominator and the put it back together as a fraction. Again, if you’re not sure you believe this hold on until the next section and we’ll take a more detailed look at this.

We can simplify this rational expression however as follows.

\[h\left( t \right) = \frac{{2{t^5}}}{{{t^2}}} + \frac{{{t^2}}}{{{t^2}}} - \frac{5}{{{t^2}}} = 2{t^3} + 1 - 5{t^{ - 2}}\]

This is a function that we can differentiate.

\[h'\left( t \right) = 6{t^2} + 10{t^{ - 3}}\]

We know that the rate of change of a function is given by the functions derivative so all we need to do is it rewrite the function (to deal with the second term) and then take the derivative.

\[f\left( x \right) = 2{x^3} + 300{x^{ - 3}} + 4\hspace{0.25in} \Rightarrow \hspace{0.25in}f'\left( x \right) = 6{x^2} - 900{x^{ - 4}} = 6{x^2} - \frac{{900}}{{{x^4}}}\]

Note that we rewrote the last term in the derivative back as a fraction. This is not something we’ve done to this point and is only being done here to help with the evaluation in the next step. It’s often easier to do the evaluation with positive exponents.

So, upon evaluating the derivative we get

\[f'\left( { - 2} \right) = 6\left( 4 \right) - \frac{{900}}{{16}} = - \frac{{129}}{4} = - 32.25\]

So, at \(x = - 2\) the derivative is negative and so the function is decreasing at \(x = - 2\).

We know that the equation of a tangent line is given by,

\[y = f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\]

So, we will need the derivative of the function (don’t forget to get rid of the radical).

\[f\left( x \right) = 4x - 8{x^{\frac{1}{2}}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}f'\left( x \right) = 4 - 4{x^{ - \frac{1}{2}}} = 4 - \frac{4}{{{x^{\frac{1}{2}}}}}\]

Again, notice that we eliminated the negative exponent in the derivative solely for the sake of the evaluation. All we need to do then is evaluate the function and the derivative at the point in question, \(x = 16\).

\[f\left( {16} \right) = 64 - 8\left( 4 \right) = 32\hspace{0.25in}\hspace{0.25in}f'\left( 16 \right) = 4 - \frac{4}{4} = 3\]

The tangent line is then,

\[y = 32 + 3\left( {x - 16} \right) = 3x - 16\]

The only way that we’ll know for sure which direction the object is moving is to have the velocity in hand. Recall that if the velocity is positive the object is moving off to the right and if the velocity is negative then the object is moving to the left.

We need the derivative in order to get the velocity of the object. The derivative, and hence the velocity, is,

\[s'\left( t \right) = 6{t^2} - 42t + 60 = 6\left( {{t^2} - 7t + 10} \right) = 6\left( {t - 2} \right)\left( {t - 5} \right)\]

The reason for factoring the derivative will be apparent shortly.

Now, we need to determine where the derivative is positive and where the derivative is negative. There are several ways to do this. The method that we tend to prefer is the following.

Since polynomials are continuous we know from the Intermediate Value Theorem that if the polynomial ever changes sign then it must have first gone through zero. So, if we knew where the derivative was zero we would know the only points where the derivative might change sign.

We can see from the factored form of the derivative that the derivative will be zero at \(t = 2\) and \(t = 5\). Let’s graph these points on a number line.

Now, we can see that these two points divide the number line into three distinct regions. In each of these regions we know that the derivative will be the same sign. Recall the derivative can only change sign at the two points that are used to divide the number line up into the regions.

Therefore, all that we need to do is to check the derivative at a test point in each region and the derivative in that region will have the same sign as the test point. Here is the number line with the test points and results shown.

Here are the intervals in which the derivative is positive and negative.

\[\begin{array}{rl}{{\mbox{positive : }}}&{ - \infty < t < 2\,\,\,\,\& \,\,\,\,5 < t < \infty }\\{{\mbox{negative : }}}&{2 < t < 5}\end{array}\]

We included negative \(t\)’s here because we could even though they may not make much sense for this problem. Once we know this we also can answer the question. The object is moving to the right and left in the following intervals.

\[\begin{array}{rl}{{\mbox{moving to the right : }}}&{ - \infty < t < 2\,\,\,\,\& \,\,\,\,5 < t < \infty }\\{{\mbox{moving to the left : }}}&{2 < t < 5}\end{array}\]