#### Recent Posts

- Reading Assignment: pp. 325-top 329
- Reading Assignment: pp. 210-211 (Examples 1-2), p. 213 (Example 4), p. 219 (Examples 2-5)
- Reading Assignment: bottom of p. 177 and top of p. 178 (proof of Sum/Difference Rule), pp. 186-187 (expls. 1-3), p. 188 (expls. 4-5)
- Reading Assignment: pp. 125-127 (start at bottom of p. 125), 133-134
- Reading Assignment: pp. 27-29, 36-38

#### Recent Comments

- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329

Example 1 on page 210 is a little confusing as it does not really explain how it got from step to step. But, for the most part, this reading if pretty self explanatory- Just derive both sides separately and plug in the point (when given).

Yes, exactly. Keeping in mind we must multiply by y’ every time we take a derivative of a y-term, we differentiate

x^2 + y^2 = 25

to get

2x + 2y*y’ = 0.

Note that book uses the notation dy/dx for the derivative, as opposed to the y’ notation that we frequently use in class.

So far the section is pretty self explanatory just more derivatives and deriving both sides. A little confused by the radical in the derivative in solution 2 on page 210. on page 219 confused on why the formula 2 and chain rule are combined?

Yeah that is confusing to me as well. But for me it relates back to how the chain rule and the rules for implicit differentiation is confusing for me also. So my question is, do you always put a y’ whenever there is another y written in the equation?

You only put a y’ when you do a derivative of a term containing y.

A few examples:

[x^2]’ = 2x (usual rules)

[y^2]’ = 2y*y’ (usual rules for the first part, but also multiply by y’ since we did a derivative of a y-term).

[x^2*y^2]’ = (x^2)(2y*y’)+(2x)(y^2)

This last example contains both x and y, so we use the product rule (F*S’+F’*S), with F=x^2, S=y^2, F’=2x, S’=2y*y’. Notice that the F*S’ part contains y’ because in it, we did a derivative of a y-term (contains derivative of S). The F’*S part does not have a y’ because in it, we did not do a derivative of a y-term (contains derivative of F, which does not contain a y).

The radical appears because they are solving x^2 + y^2 = 25 for y.

This gives y^2 = 25 – x^2. Taking the square root of both sides, and remembering to put the +/- on the right, we get

y = +/- sqrt(25 – x^2).

Now you can find y’ using the “normal” rules. Since y contains a square root, first rewrite as a power, y = +/- (25 – x^2)^{1/2}, and then use the chain rule.

Formula 2 is that [ln(x)]’ = 1/x. If we want [ln(g(x))]’ instead, we must use Formula 2 and the chain rule (we did this in class). In this case, inner = g(x) and outer = ln(x). So inner’ = g'(x) and outer’ = 1/x. So using the chain rule (outer'(inner)*inner’),

[ln(g(x))]’ = 1/g(x) * g'(x) = g'(x)/g(x).

For example 1 on page 210, what is the difference between the notations d/dx, d/dy, and dy/dx? The method in class an this example make sense, but I’m just wondering what those notations mean and how they’re introduced into the problem. Example 4 on pg 213 makes sense, d/dx notation intimidates me but the y’ is easy to solve for and make sense of. All the rules for logarithmic functions make sense on pg 219. However, for example 4, the 1/xlna was multiplied by the derivative of “x” (a function) but the rule didn’t add that in on pg 218. When don’t we multiply by the derivative of “x”? Or is the rule implying you multiply by the derivative of x, and since it is 1 they left it out?

As far as derivative notation, there is a lot. They all mean the same thing, but are just different ways of writing it. Here are some examples for the function f(x)=x^2 (which could also be written y=x^2):

f'(x) = 2x

y’ = 2x

dy/dx = 2x

d/dx[x^2] = 2x

[x^2]’ = 2x.

You also mention d/dy notation. The meaning is the same as d/dx, except you are treating y as the independent variable, rather than x.

So for example,

d/dy[x^2] = 0, since x^2 contains no y’s (if y changes, x^2 doesn’t, so has derivative 0). But

d/dy[y^2] = 2y.

In example 4 on page 219, the part you ask about requires using the 1/xlna rule along with the chain rule.

d/dx[log_a(x)] = 1/xlna

Use this with the chain rule to get the last rule we mentioned in class:

d/dx[log_a(g(x))] = 1/g(x)lna * g'(x).

You can think of the case where g(x)=x as a special case of this last rule. If the inner function is simply “x”, then it’s derivative is 1, so the g'(x) part is typically left out.

Implicit differentiation confuses me, I think it’s because I don’t understand the chain rule/product rule and they sort of build on each other. I understand that we are trying to solve for y prime. SO going back to the chain rule, I understand that as an individual looking for outer prime and inserting inner while multiplying that by inner prime. But I get very confused when I have to use that and the product/quotient rule. If I have to use the product rule I do the first and second but where do I incorporate the chain rule. Back to implicit, the whole scheme has to do with get y prime to one side and solving for that. But I don’t understand the problem when we have to use the chain rule. y double prime makes a lot of sense but I can’t even get to that part because I’m stuck on for the derivative of y. Logarithm functions kill me! when did I example 2 on page 219 get 1/sinx? Can we go over using the chain rule along with other rules? I think once I get the grasp of that then I’ll be okay.

There are cases where the chain rule is used within the product rule, and other cases where the product rule is used within the chain rule.

For example, if y = (x+5)^10 * (x-1)^8, you would start with the product rule

F = (x+5)^10

S = (x-1)^8

To find F’ and S’, however, you need to use the chain rule (both of them have the type “function to a power”).

If y = e^{x*sinx}, you would start with the chain rule since the function has the type “e raised to a function”. As part of the chain rule, you need the derivative of inner = x*sinx. Since this is a product of two functions, to find this you need the product rule, where F=x and S=sinx. So inner’ = F*S’ + F’*S.

When dealing with implicit differentiation, think of identifying terms as one of three types: x-terms contain only x’s and we use the same rules we learned in sections 3.1-3.5; y-terms contain only y’s and we use the same rules we learned in sections 3.1-3.5, but multiply by y’ every time a derivative of y is done; mixed terms contain both x’s and y’s, and we generally must used the product or quotient rule or chain rule, depending on the form. Handling these terms correctly requires a lot of practice.

For example 2 on page 219, 1/sinx is outer'(inner). Since inner=sinx and outer=lnx, outer’=1/x. So outer'(inner) = 1/sinx (note that this is the inner function plugged into the derivative of the outer function; so sinx plugged into 1/x).

I think it’s really interesting how implicit differentiation helps to find the derivative much faster than by solving for y.

For example 5 on page 219, I’m a little bit confused on how they got 1/x+1 in the numerator as the first step. Are they using the chain rule? Or are they using the derivative of ln?

Remember the rule for d/dx[ln(g(x))] is 1/g(x) * g'(x). In example 5, the inner function is g(x)=(x+1)/sqrt(x-2). So

1/g(x) = 1/(x+1)/sqrt(x-2).

So actually think of the numerator as 1 and the denominator as g(x) = (x+1)/sqrt(x-2).

For example 1 on page 210, I fully understand parts a and b. For example 2 on page 211, I understand everything up until part c. The explanation gets a bit confusing and I lose them there. For example 4 on page 213, I know this isn’t as important, but why do they use d/dx to show some derivatives and ‘ to show others? For all examples on page 219, I donâ€™t understand what the log’s and the ln’s do to the problems.

For example 2c, recall that a horizontal line has slope 0, and the slope of the tangent line is the derivative. So for a tangent line to be horizontal, the derivative needs to be 0.

Since y’ is a fraction, it is 0 only when the top is 0 (and the bottom isn’t). So y’ = 0 if 2y-x^2=0, which means y=.5x^2. Since x and y need to satisfy the original equation, x^3+6^3 = 6xy, they substitute y=.5x^2 into this to solve for x.

As far as the notation, the d/dx() notation means “the derivative of what is in the parentheses”. They often use this to break up the problem into pieces.

In example 4, to find the derivative of -x^3/y^3 (which can be written as d/dx(-x^3/y^3)), we need to use the quotient rule, which starts out as B*T’….

Notice that they write this as y^3(d/dx)(x^3), which is just the bottom times the derivative of the top. In the next line, they actually take the derivative of the top and write this as y^3*3x^2.

On page 219, the log’s and ln’s are part of the function whose derivative you are trying to find. To find these, you need to use rules for derivatives of log functions (page 218, or we called these rule 12 in class).

On page 210-211 I understood examples both examples for the most part but example 1 confused me a bit. I don’t understand how they got dy/dx= -x/y. I understood that they substituted the coordinates (3,4) into the x and y value but I don’t get how they got a negative in the front of it. And it is the same with example 4 on page 213 I understand them doing the quotient rule but I do not understand how they got y prime= -x^3/y^3. Why is it negative? But example 2-5 219 I understood with no problem.

First, dy/dx means the same as y’ (both means the derivative of y, treated as a function of x).

Near the middle of the page, they have

2x + 2y*y’ = 0.

If we solve for y’, we get

2y*y’ = -2x

y’ = -2x/2y

y’ = -x/y.

On page 213, same idea. The negative comes from the fact that we are subtracting the term that does not have y’ in it over to the right.

I could understand everything that was in the reading about y’ but when it comes to logarithmic functions and finding the derivatives of those I get a bit confused. I don’t really know the natural log can be canceled within functions and why exactly ln turns into log. I would like some clarification on examples 2-4 on page 219.

In examples 2-4, the natural log isn’t really being “cancelled”, nor is a natural log turning into a log. The rules for derivatives of logarithms are being used.

For example, one rule is

d/dx[ln(g(x))] = g'(x)/g(x). This rule says if we want the derivative of the natural log of a function, we should do the derivative of the inner function divided by itself.

So in example 2, the inner function in d/dx[ln(sinx))] is sinx, so this rule says the derivative should be the derivative of sinx over itself, i.e., cosx/sinx (which is cotx since it’s the reciprocal of tanx).

Similar idea in example 3, except the rule is

d/dx[log_a(g(x))] = g'(x)/g(x)lna. This rule says if we want the derivative of the log of a function (not base e), we should do the derivative of the inner function divided by the product of itself with the natural logarithm of the base.

So in example 3, the inner function in d/dx[log_10(2+sinx))] is 2+sinx, so this rule says the derivative should be the derivative of 2+sinx over the product of itself with ln(10), i.e.,

cosx

————–

(2+sinx)ln(10)

Implicit differentiation is a concept that I find easy to understand. Both of the examples provided on page 210 and 211 are easy to follow and provide for a better understanding. Finding the second derivative of something can always be more confusing but the example provided on page 213 really makes it clear in how to do so. Page 219 is more confusing to me, the examples provided make it a little more clear. Overall I find implicit differentiation to be an easier topic and just need to work on the logarithmic functions.

I only have one question, it’s more of a general question but will we ever have more than one log function in an equation? Also, if the question has a log base 10x, would we have to take the derivative of the 10x after weDo the 1/ln10x? It’s just something I thought of while looking at example 4.

Generally there will only be one logarithm in a function. It’s possible to have more than one, and a more challenging homework problem may have more than one (for example # 15 in 3.6 is F(s)=ln(ln(s)), which means plug s into natural log to get ln(s), then plug the result into the natural log function again to get ln(ln(s))), but it’s not the “norm”.

The base of a logarithm is always a number (without a variable), so in example 4, the base is just 10 (not 10x). Remember that log_10(x) means you plug x into the log base 10 function, which really means what power should 10 be raised to to produce x.

So they are using the rule d/dx[log_10(x)]=1/(x*ln(10)) in example 4, along with the chain rule.