Feelings on Exam 1, and the newest rules for Derivatives

Exam 1 went well. Oh… except for the fact that I am 100% positive I missed the first question. All I remember was I was supposed to find the limit of x at 3, from the left, and I forgot that these little symbols

|x or whatever equation happens to be inside| mean only for positive numbers, so when you get the answer -2, the real answer should have been just 2.

I can’t remember some of the simpler rules from algebra and my previous math courses, and that really annoys me. I literally realized it the minute I walked out of the test, but it’s not like you can run back in and demand a redo.

So now I have to bank that knowledge away and never let it escape me again.

On to what we learned in class today…

The Chain Rule:

If G(x) = F( U(x) ) then G'(x) = F’ (U(x)) * U’

That’s a lot of confusing letters, so I’ll break it down for you further..

If $G(x) = (3x + 5)^8$ then $G'(x) = 8(3x+5)^7 * 3$

If you’re still confused I will show you what I did

G(x) = F( U(x) ) = G(x) = $(3x+5)^8$

Now we have to figure out what F is and what U(x) is.

F is $x^8$ and U(x) fills in for what would’ve been x in F(x). So instead of x, we put U(x) which in this case is $(3x + 5)$

Now that we have our two functions we can derive F(x) into $F'(x) =8x^7$

and $U'(x) = 3$

Thus G'(x) = F'(U(x))*U’ = $G'(x) = 8(3x+5)^7*3$ and simplified further becomes $G'(x) = 24(3x+5)^7$

We did some practice equations using the chain rule in class today and I found there were a few people in class who were TOTALLY confused behind me on some of the more difficult problems, so I thought I would put in 2 of the harder equations and show how I solved for them.

Using the Chain Rule with a Square Root:

Take the equation $y=\sqrt{x^2+3x+1}$ find $\frac{dy}{dx}$

So to make this a little more broken down, first I am going to identify which parts of the equation I need to work with.

Let’s start with finding our “F(x)” – in this case it would be $\sqrt{x}$

In order to find the derivative, we have to make the square root into an exponential equation, so $F'(x) = \frac{1}{2}x^{\frac{-1}{2}}$ If you don’t know how I got that, look at some of my older posts on square roots and derivatives.

Now we need to figure out what our “U(x)” is, in this case it’s $(x^2+3x+1)$

U'(x) is $2x+3$ so now we can plug this in to our formula and we get this:

$\frac{dy}{dx}= \frac{1}{2}(x^2+3x+1)^{\frac{-1}{2}}*(2x+3)$

but we have to simplify and get it back in square root form, so because the exponent of U(x) is negative, we’ve got to put it on the bottom, and the solution looks like this:

$\dfrac{dy}{dx}=\dfrac{2x+3}{2 \sqrt{x^2+3x+1}}$

Easy!

Using the Chain rule with a Fractional Function:

Then I came to a fraction equation that looked like this:

$s(t) = \frac{8}{t^3+1}$

Momentarily my brain froze.

However, after a few minutes… the answer came to me!

What? I’m trying to keep this entertaining.

In order to do this I had to create a negative exponent, so now my equation looks like this:

$s(t) = 8(t^3+1)^{-1}$

Now you can easily pick out F(x) and U(x)

$F(x) = 8x^{-1}$

$F'(x) = -8x^{-2}$

$U(x) = (t^3+1)$

$U'(x) = 3t^2$

Now you can plug these into the equation and you get this:

$s'(t) = -8(t^3+1)^{-2}*3t^2$

$s'(t) = -24t^2(t^3+1)^{-2}$

and you have to put it back in fractional form so the final answer is:

$s'(t) = \dfrac{-24t^2}{(t^3+1}^2$

Using the Chain Rule with the Product Rule:

Finally we came to a difficult equation, and the teacher had to walk us through using the product rule with the chain rule. Here’s what it looked like:

$y=x^3(x^2+1)^5$

In order to get to the chain rule, we have to use the product rule, $y'=ab'+ba'$

so here, we have to find a, and b, and the derivatives of both.

$a=x^3$

$a'=3x^2$

$b= (x^2+1)^5$

To find the derivative of b, we have to use the chain rule. So $f(x) = x^5$ and $U(x) = x^2+1$ So $F'(x)=5x^4$ and $U'(x) = 2x$ Plug it in and you get:

$b'= 5(x^2+1)^4 * 2x$

Now we can plug in both to make the equation y’=ab’+ba’

$y'= x^3*10x(x^2+1)^4+(x^2+1)^5*3x^2$

This can be simplified into $y'= 10x^4(x^2+1)^4+3x^2(x^2+1)^5$

and further simplified by factoring out somethings.. we have

$x^2(x^2+1)^4$ which can be factored out and that leaves us with:

$10x^2 + 3(x^2+1) = 10x^2+3x^2+3 = 13x^2+3$ which we can now recombine with what we factored out for a more cohesive equation:

$y'=x^2(x^2+1)^4(13x^2+3)$

I hope this helps.. if it doesn’t, please feel free to ask me questions. I tried to make it comprehensible but it’s hard when the teacher is so clearly comprehensible to me so I am not always sure how to make it more so.

Finally, I solved the last equation on the front page of our worksheet, which looked scary, but it wasn’t so bad once we incorporate all the stuff we’ve learned.

Using the Chain Rule with Quotient Rule

$y=\dfrac{(x^3+5)^7}{8x-3}$

First what I want to do is turn this into a multiplication equation and solve for the product rule, however I wanted to try this using the quotient rule. Here goes nothing.

Quotient rule: $y'= \dfrac{bt'-tb'}{b^2}$

b=8x-3 and b’=8

$t=(x^3+5)^7$ in order to get t’ we have to use the chain rule

$t'= 7(x^3+5)^6*2x^2$

Now that we have everything, we can plug it in to the quotient rule:

$y'= \dfrac{(8x-8)*21x^2(x^3+5)^6-7(x^3+5)^7(8)}{(8x-3)^2}$

That’s a big one.

you can factor some stuff out though… $(x^3+5)^6$ and when you do that you can simplify what’s remaining $(8x-3)(21x^3)-56(x^3+5)$ into $168x^3-63x^2-56x^3-280$