Question

Differentiate the following functions using the rules
y=kx^n --> dy/dx = knx^n-1
22. y= 1/x
23. y square root of x
26. f(x)=5/x^3
27. y=2/square root of x

Answer 1

These are some I am having trouble with and have gotten wrong, please can someone show me the steps and maybe explain the rules behind it a little?

Answer 2

They would like you to apply Power Rule to each of these?
Ah nice :) Good practice problems.

Answer 3

This requires that you remember some of your exponent rules.

Answer 4

\[\large\rm \frac{1}{x^1}=x^{-1}\]Ok with this one?

Answer 5

Applying your power rule is going to feel a little strange since our power is negative.
Instead of decreasing,
the number will increase in the negative direction.\[\large\rm y=x^{-1}\qquad\to\qquad y'=-1\cdot x^{-1-1}\]So,\[\large\rm y'=-x^{-2}\]Questions? :o

Answer 6

Ok, no that makes sense, the answer given is -1/x^2, is there a specific reason that its in this form?

Answer 7

They applied the exponent rule back in the other direction.
Negative powers look kind of ugly,
it's more appropriate to give our final answer in fraction form usually.
Not a big deal though.

Here is the general exponent rule though:\[\large\rm x^{-a}=\frac{1}{x^a}\]That's what it would look applying the rule in that direction.\[\large\rm -x^{-2}=-\frac{1}{x^2}\]Exponent changes from negative to positive, and the fraction gets flipped.

Answer 8

Ah I see, ok. That one is good now

Answer 9

For 23, I thought it was \[\frac{ 1 }{ 2 }x\]

Answer 10

Mmmm ok that's headed in the right direction at least.
The exponent is off though.

Answer 11

So we rewrite our root as a rational exponent,\[\large\rm y=\sqrt x=x^{1/2}\]And apply power rule,\[\large\rm y=\frac{1}{2}x^{\frac{1}{2}-1}\]

Answer 12

So what is your new power going to be?

Answer 13

\[\frac{ 1 }{ 2 }x ^{-\frac{ 1 }{ 2 }}\]

Answer 14

Good good good.
And you can apply your exponent rules again to make it look a little prettier.

Answer 15

\[\large\rm \frac{1}{2}\cdot x^{-1/2}\quad=\quad \frac{1}{2}\cdot\frac{1}{x^{1/2}}\quad=\quad \frac{1}{2}\cdot\frac{1}{\sqrt x}\quad=\quad\frac{1}{2\sqrt x}\]

Answer 16

But hey, before we move onto the next one.
`Remember this one`.

Square root comes up over and over and over.
Stick this in your back pocket and never use power rule on it again.

Derivative of square root is `1 over 2 square roots`.

Answer 17

This derivative is as valuable as ln(x).
So just put it to memory :)

Answer 18

Ok, noted. Will keep that in mind, thanks!

Answer 19

\[\large\rm f(x)=\frac{5}{x^3}\]Hmm any ideas? :d

Answer 20

I got -15x^-4....

Answer 21

Ooo very nice.
And maybe we make it look pretty as a final step: \(\large\rm -\dfrac{15}{x^4}\)

Answer 22

Oh ok. Just wasnt thinking on that one :)

Answer 23

For 27,\[\large\rm y=\frac{2}{\sqrt x}\]the shortcut I mentioned in 23.. we actually can't make good use of that here,
so we'll just turn the root into a number and do the same ole jazz as before.
The new power is going to look a little awkward though.

Answer 24

sooo \[2x ^{\frac{ 1 }{ 2 }}\]
Then
\[x^-{\frac{ 1 }{ 2 }}\]?

Answer 25

Is the "sooo" your function y,
and "then" is your derivative?

Answer 26

Well first I was trying to write it differently and the I applied the power rule, I dont think I did it properly though so I was unsure.

Answer 27

Gotta fix the exponent on the original function, it didn't work out correctly.\[\large\rm y=\frac{2}{\sqrt x}=\frac{2}{x^{1/2}}=2x^{-1/2}\]This is what you should have before differentiating ^

Answer 28

So next it would be \[x^\frac{ -3 }{ 2 }\]

Answer 29

oh negative x not positive because I multiplied by -1/2 right?

Answer 30

\[\large\rm y'=-\frac{1}{2}\cdot2x^{-3/2}\]Negative, good catch!

Answer 31

So ya, these aren't too bad right?
Have you learned any of the other shortcut rules yet?

Answer 32

Yes not to bad! Just need to focus as I am good at making very small mistakes that I dont notice (practice is the answer to that I guess). I am not quite sure what you mean?

Answer 33

Power Rule
Product Rule
Quotient Rule
`Chain Rule`

These are the 4 handy shortcut rules for dealing with differentiation.
When Chain Rule comes up, really really pay attention.
It's by far the trickiest of these rules and will take the most practice to get comfortable with.

Answer 34

I think I recognize all of them except the chain rule, but I will start practicing all of them as extra. Didnt really know the names and how many of them there were before

Answer 35

Thank you so much for your help! I really appreciate it!

Answer 36

Np! ୧ʕ•̀ᴥ•́ʔ୨