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# Calculus Derivatives and tangent equations

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788
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+1314
Online test questions, you don't have to give me the answer as I do want to learn the process and be able to do it, but I suspect that the way the test is set up, that investigation and getting help is part of the process to encourage learning of difficult material. So answers are not necessary but mighty helpful since, I can see what is wrong, thus feel free to put new variables in. Note there is a few questions here, challenging for me, and since I can not follow the derivative rules to well, or if I do the answers are not put it in the correct form etc. The last answers I got were input right so thank you and that shows that if the process is right the system will accept the answers.

1
http://snag.gy/w8UHZ.jpg

2
http://snag.gy/FrQ15.jpg

3
http://snag.gy/sCKWg.jpg

4
http://snag.gy/eE7hY.jpg
I really would like an explanation of this question and answer, or if there was a another answer to arrive at through a different method and what that method would be. thanks.

I appreciate the help.
Apr 5, 2014

#6
+109783
+5

The chain rule is very basic and I do not think that it is expained to students very well.

Do you know how you can cancel in fractions

eg

speed*time = distance

because

$$\frac{6km}{hour}\times\frac{7hour}{1}=42km$$  Because the hours cancel out!

With calulus the d in front of the x and the y etc just stands for difference.  You can cancel out the units just like I did in the above example.

So for instance

$$\frac{dy}{du}\times \frac{du}{dx}\times \frac{dx}{dt}=\frac{dy}{dt}$$     The du's and dx's cancel out.

That is the chain rule!

Jun 2, 2014

#1
+2353
+5
Stu:

Online test questions, you don't have to give me the answer as I do want to learn the process and be able to do it, but I suspect that the way the test is set up, that investigation and getting help is part of the process to encourage learning of difficult material. So answers are not necessary but mighty helpful since, I can see what is wrong, thus feel free to put new variables in. Note there is a few questions here, challenging for me, and since I can not follow the derivative rules to well, or if I do the answers are not put it in the correct form etc. The last answers I got were input right so thank you and that shows that if the process is right the system will accept the answers.

1
http://snag.gy/w8UHZ.jpg

2
http://snag.gy/FrQ15.jpg

3
http://snag.gy/sCKWg.jpg

4
http://snag.gy/eE7hY.jpg
I really would like an explanation of this question and answer, or if there was a another answer to arrive at through a different method and what that method would be. thanks.

I appreciate the help.

Okay, let's see what I can do here,
I think I'll give you important hints for the first three and show you the whole solution to the fourth one to give you an idea.

Also I think it might help you to look at this site; http://www.mathsisfun.com/calculus/derivatives-rules.html
This website has lot's of simple explanations on various mathematical subjects and this section pretty much summarizes the rules for derivatives.

Now, for the first one, you should look into the product rule. The idea is that f(x) = g(x)*h(x) then f'(x) = g'(x)*h(x) + g(x)*h'(x)

So for example if g(x) = x 2 and h(x) = e x so f(x) = x 2*e x

then f'(x) = 2xe x + x 2e x = (2x+x 2)e x

For the second one, look at the derivatives of trigonometric functions (and learn them by heart). So, f(x) = sin(x) => f'(x) = cos(x) and g(x) = cos(x) => g'(x) = -sin(x)

Also have a look at the chain rule to see what to do with cos(pi*x).

For the third one,

Same as the second one, look at the derivatives of trigonometric functions and the chain rule. That should do the trick

I'll do the fourth one in a seperate post.
Apr 5, 2014
#2
+30386
+5
Here's one possible approach to the first part of the fourth problem. Notice that I've used the df/dx notation for derivatives rather than the f' notation, but it means the same thing.

f(x) = (2x+1) 9(3x-1) 7

Stuproblem.PNG

The steps might seem long-winded, but if you take it step by step it shouldn't be too bad.

There are other ways to approach this (perhaps Reinout will use a different one.)
Apr 5, 2014
#3
+2353
+5
Okay, so f(x) = (2x+1) 9(3x-1) 7.

Now we have two functions so we want to apply the product rule.

The product rule works for functions in the following way f(x) = g(x) * h(x)

in this case g(x) = (2x+1) 9 and h(x) = (3x-1) 7

Since we know the product rule states f'(x) = g'(x)*h(x) + g(x) * h'(x)

we have to calculate the derivatives of g(x) and h(x).

For both we need the chain rule which states f(x) = g(h(x)) => f'(x) = g'(h(x))*h'(x).

That equation might be an appaling sight so let me attempt to state it to you in simpler words.

The derivative of a function 2 within a function 1 is the derivative of the function 1 leaving the function 2 as it is multiplied by the derivative of the function 2

Since this still might seem a little fuzzy let me give you an example with the functions we have g(x) = (2x+1) 9

We could again rewrite this to g(x) = (u(x)) 9. where u(x) = 2x+1 (so u'(x) = 2) then g'(x) = 9*u(x) 8*u'(x) = 9*(2x+1) 8*2 = 18(2x+1) 8

Now, let me continue with the rest of the original question.

We just saw that g'(x) = 18(2x+1) 8

Similarly we can calculate (again using the chain rule) h'(x) = 7*(3x-1) 6*3 = 21*(3x-1) 6

Then for f'(x), we can fill in the blanks.

f'(x) = g'(x)h(x) + g(x)h'(x) = 18(2x+1) 8* (3x-1) 7 + (2x+1) 9*21(3x-1) 6

We can simplify this to 18(2x+1) 8*(3x-1) 6*(3x-1) + (2x+1) * (2x+1) 8 * 21(3x-1) 6

I did this so both terms have (2x+1) 8*(3x-1) 6 in common.

This means I can now simply the equation to (18(3x-1)+21(2x+1)) * (2x+1) 8*(3x-1) 6

Note that: 18(3x-1)+21(2x+1) = 54x-18+42x+21 = 96x+3 = 3(32x+1)

So we calculated that f'(x) = 3(32x+1)(2x+1) 8(3x-1) 6

f"(x) works in quite a similar way, where we can again use f(x) = g(x)h(x) only this time g(x) also is a multiplication of two functions so for g'(x) you'd also need to use the product rule.

Must be a case of productruleception

See whether you can do that one yourself.

Reinout

edit: just saw Alan did the same thing in the mean time
Apr 5, 2014
#4
+111396
+7
1. y = (t^2 + 5)e^t find dy/dt (or y')
There are a couple of different ways we could do this one, but let's call the expression inside the parenthesis f and the other part g.
So, we have y = fg
And, using the product rule, we have y' = f 'g + f g '
so f ' = 2t and g ' = e^t
And putting this together gives us
y' = (2t)e^t + (t^2 + 5)e^t ..... And simplifying we have
y' = e^t(t^2 + 2t + 5)

2. R(x) = 11 - 2cos(Π x)
Find R ' (x)
OK......Let's just concentrate on the "cos(Π x) " part....we'll go back and finish the rest in a sec
Note, that we can use the chain rule here. Call cos(Π x) the "outside" function and call Π x the "inside" function,
So, the derivative of the "outside function" is just -sin(Π x) and the derivative of the "inside" function is just Π
Remembering that the derivative of the constant 11 is just zero, we have
R ' (x) = -2(-sin(Π x) * Π).....And simplifying this, we have
R ' (x) = 2Πsin(Π x)

3. w = sin(8e^t)
Find w'
This is pretty much like the last one. We're going to use the chain rule here, too. Call sin(8e^t) the "outside" function and call 8e^t the "inside" function.
Then the chain rule says to take the derivative of the "outside" times the derivative of the "inside."
So the derivative of sin(8e^t) is just cos(8e^t) and the derivative of 8e^t is just itself.
And putting this together we get
w' = cos(8e^t) * 8e^t or just
8e^t cos(8e^t)

4. f(x) = y = (2x + 1)^9 (3x - 1)^7 .....Find the first and second derivatives.
This one is more "tedious" than difficult. It combines both the chain and product rules (plus a little simplifying).
As before, let's call (2x + 1)^9, "f" and let's call (3x - 1)^7, "g."
Then, y = fg and y' = f 'g + f g '
So, to calculate f ' , we'll need to employ the chain rule. Let's call (2x + 1)^9 the "outside" function and 2x + 1 the "inside" function.
The derivative of the outside is 9(2x + 1)^8 and the derivative of the inside is just 2. And multiplying these together we get 18(2x + 1)^8.
Similarly, we'll need the chain rule again to find g'. Let (3x - 1)^7 be the "outside" function and let 3x - 1 be the "inside" function.
The derivative of the outside is 7(3x - 1)^6 and the derivative of the inside is just 3. And multiplying these together we get 21(3x - 1)^6.
So, putting all this together, we get.
y' = f 'g + f g '
y' = 18(2x + 1)^8 (3x - 1)^7 + (2x + 1)^9 (21(3x - 1)^6)
Let's factor out the lowest powers on the "(2x + 1)" and "(3x - 1)" terms.
This gives (2x + 1)^8 * (3x - 1) ^6 * [18(3x - 1) + 21(2x + 1)]
And simplifying inside the brackets we have (2x + 1)^8 * (3x - 1) ^6 * [96x + 3]
And factoring a "3" out of [96x + 3], we have
3*(2x + 1)^8 * (3x - 1) ^6 * (32x + 1)

To find y '', we just extend the product rule over one more function, i.e.
y' = fgh
y'' = f' gh + fg' h + fgh'
To make things easier, I'm going to put our last answer back into this form....... (2x + 1)^8 * (3x - 1) ^6 * [96x + 3], where "f" is (2x + 1)^8, "g" is (3x - 1) ^6, and "h" is [96x + 3]
To save some time, (And because I think you see how to apply the chain rule now), f ' = 8(2x + 1)^7 * (2) = 16(2x + 1)^7 and g ' = 6(3x - 1) ^5 *(3) = 18(3x - 1) ^5 and h' = 96
Sp...let's put everything together and simplify...we have
y'' = f' gh + fg' h + fgh'
y''= (16(2x + 1)^7) * (3x - 1) ^6 * [96x + 3] + (2x + 1)^8 * 18(3x - 1) ^5 * [96x + 3] + (2x + 1)^8 * (3x - 1) ^6 * [96]
So, (as before), factoring out the lowest powers of the (2x + 1 ) and (3x -1) terms, we have
(2x + 1)^7 * (3x - 1)^5 [(16 (3x - 1)* (96x + 3) +(18) (2x +1) (96x +3) + (2x + 1) (3x -1 ) (96)] =
(2x + 1)^7 * (3x - 1)^5 [ 16 (288x^2 - 87x - 3) + 18 (192x^2 + 102x + 3) + 96(6x^2+ x -1)]
(2x + 1)^7 * (3x - 1)^5 [8640x^2 + 540x - 90]
Finally, we can factor the last term as 10 * (864x^2 + 54X - 9) = 10 * 3 * ( 288x^2 + 18x - 3) = 10 * 3 * 3 * (96x^2 + 6x - 1) = 90 (96x^2 + 6x - 1)
So we get
90 *(2x + 1)^7 * (3x - 1)^5 * (96x^2 + 6x - 1)

Whew!!! That was a long one, huh??

I hope this helps.....
Apr 5, 2014
#5
+1314
0

you mentioned the chain rule a few times, is that, product rule states f'(x) = g'(x)*h(x) + g(x) * h'(x) the same as uv' + vu'. Looks like it. Then what is To find y '', we just extend the product rule over one more function, i.e.
y' = fgh
y'' = f' gh + fg' h + fgh' , and why is it applied to the second derrivative?

Thanks

Jun 1, 2014
#6
+109783
+5

The chain rule is very basic and I do not think that it is expained to students very well.

Do you know how you can cancel in fractions

eg

speed*time = distance

because

$$\frac{6km}{hour}\times\frac{7hour}{1}=42km$$  Because the hours cancel out!

With calulus the d in front of the x and the y etc just stands for difference.  You can cancel out the units just like I did in the above example.

So for instance

$$\frac{dy}{du}\times \frac{du}{dx}\times \frac{dx}{dt}=\frac{dy}{dt}$$     The du's and dx's cancel out.

That is the chain rule!

Melody Jun 2, 2014