Question

use logarithmic differentiation to find the derivative of the function. y = (sqrt(xe^x^2))(x^2+1)^7

Answer 1

\[\large y=\sqrt{x e^{x^2}}(x^2+1)^7\]
Is this what our function looks like?

Answer 2

yea

Answer 3

Taking the log of both sides,\[\large \ln y=\ln\left[\sqrt{x e^{x^2}}(x^2+1)^7\right]\]We can use rules of logarithms to break apart the right side.
Here is one rule we'll be using:
\(\large \log(ab)=\log a + \log b\)

Answer 4

Which allows us to split it up like this:
\[\large \ln y=\ln\left[\sqrt{x e^{x^2}}\right]+\ln\left[(x^2+1)^7\right]\]

Answer 5

This first term we can umm, rewrite the square root as a rational expression:\[\large \ln\left[\sqrt{x e^{x^2}}\right] \qquad=\qquad \ln\left[({x e^{x^2}})^{1/2}\right]\]Applying the exponent to everything inside the brackets:\[\large = \qquad \ln\left[x^{1/2}e^{x^2/2}\right]\]Then we can use our rule from earlier to break this down further:\[\large = \qquad \ln\left[x^{1/2}\right]+\ln\left[e^{x^2/2}\right]\]

Understand what's going on here? :o
Logarithmic differentitation can be a pain in the butt.
Doing this the normal way would've been better. But they told us to use this process :P

Answer 6

lol. i am stating to understand it better. breaking it down like this is helping alot

Answer 7

So we're left with this so far:\[\large \ln y= \qquad \ln\left[x^{1/2}\right]+\ln\left[e^{x^2/2}\right]+\ln\left[(x^2+1)^7\right]\]

We want to apply another rule of exponents:
\(\large \log(a^b)=b\cdot\log(a)\)

Applying this to our first term gives us:\[\large \ln\left[x^{1/2}\right] \qquad=\qquad \frac{1}{2}\ln x\]

Answer 8

Woah I dunno why that first line is bold like that :o lol

Answer 9

lol no worries

Answer 10

\[\large \ln\left[(e)^{x^2/2}\right] \qquad=\qquad \frac{x^2}{2}\ln\left[e\right]\]
Do you know what the natural log of e simplifies down to? :o

Answer 11

do u mean the derivative of ln[e]

Answer 12

No, just the value of ln e.

Answer 13

If you don't remember, throw it into your calculator :3

Answer 14

its 1

Answer 15

Ok cool, so that simplifies our middle term to:\[\large \frac{x^2}{2}\ln\left[e\right] \qquad=\qquad \frac{x^2}{2}\]

Answer 16

makes sense

Answer 17

For our last term, we'll apply this second rule of logs again:\[\large \ln\left[(x^2+1)^7\right] \qquad=\qquad 7\ln\left[x^2+1\right]\]

Answer 18

So this is what we have so far:\[\large \ln y= \qquad \frac{1}{2}\ln x+\frac{x^2}{2}+7\ln\left[x^2+1\right]\]

All we've done so far is applied a bunch of rules to make the problem easier to differentiate. Now that it's easier to work with, we'll apply the derivative from here.

Answer 19

alright

Answer 20

Are you comfortable with taking the derivative of logs?
Do you know what the left side gives us when we differentiate?

Answer 21

1/y

Answer 22

Good.
Since we're taking the derivative of y, with respect to x, we'll also get a y' when we take the derivative of that term.

\[\large \left(\ln y\right)' \qquad=\qquad \frac{1}{y}y'\]

Answer 23

\[\large \frac{1}{y}y'= \qquad \left(\frac{1}{2}\ln x+\frac{x^2}{2}+7\ln\left[x^2+1\right]\right)'\]

How bout the right side, can you figure those derivatives out? :D

Answer 24

im not exactly sure

Answer 25

Deal with each term individually :o

Answer 26

For the first term, ignore the 1/2. It won't affect the differentiation process.
We'll just multiply the derivative of lnx by 1/2.

Answer 27

would it be 1/x

Answer 28

\[\large \frac{1}{y}y'= \qquad \frac{1}{2}\cdot\frac{1}{x}+\left(\frac{x^2}{2}+7\ln\left[x^2+1\right]\right)'\]Ok good, that takes care of the first term.

Answer 29

would its then be 1/2*x^2/2

Answer 30

Hmmm

Answer 31

For this next term you're taking the derivative of \(\large \dfrac{x^2}{2}\)
Which we could write this way before taking the derivative \(\large \dfrac{1}{2}x^2\)

From there, just apply the power rule.

Answer 32

would it just be x

Answer 33

\[\large \frac{1}{y}y'= \qquad \frac{1}{2x}+x+\left(7\ln\left[x^2+1\right]\right)'\]
So the 2 came down, cancelling with the 1/2?
Mm yah that sounds right.

Answer 34

kk

Answer 35

how would you do 7lnx^2+1

Answer 36

We can pull the 7 outside, it's a constant it won't affect this process.\[\large \left(7\ln\left[x^2+1\right]\right)' \qquad=\qquad 7\left(\ln\left[x^2+1\right]\right)'\]

Mmm ok this might be a little confusing at first.
You'll need to get comfortable with your log derivative.

You determined earlier that the derivative of \(\large \ln (x)\) is \(\large \dfrac{1}{(x)}\).

The same idea follows here: the derivative of \(\large \ln\left[x^2+1\right]\) will give us \(\large \dfrac{1}{\left[x^2+1\right]}\).

The entire contents of the log will go into the denominator there.
But we also have to apply the chain rule, and multiply by the derivative of the inner function.\[\large \dfrac{1}{\left[x^2+1\right]}\left[x^2+1\right]' \qquad=\qquad \dfrac{1}{\left[x^2+1\right]}\left[2x\right]\]

Answer 37

Oh and then of course we need that 7 from before :o\[\large 7\dfrac{1}{\left[x^2+1\right]}\left[2x\right]\]

Answer 38

makes sense. is there more that needs to be done to this?

Answer 39

We have a y in the denominator on the left, so we need to deal with that.
Because our goal is to write our problem as y'=stuff.

\[\large \frac{1}{y}y' \qquad=\qquad \frac{1}{2x}+x+7\frac{1}{x^2+1}(2x)\]Multiplying both sides by y gives us,\[\large y' \qquad=\qquad \color{royalblue}{y}\left[\frac{1}{2x}+x+7\frac{1}{x^2+1}(2x)\right]\]

From here, we would like our answer to be written in terms of `x`.
That means: only x's on the right side.

Going back to our original equation:
\[\large \color{royalblue}{y=\sqrt{x e^{x^2}}(x^2+1)^7}\]

Let's plug this in for the y that's showing up in our solution.\[\large y' \qquad=\qquad \color{royalblue}{\sqrt{x e^{x^2}}(x^2+1)^7}\left[\frac{1}{2x}+x+7\frac{1}{x^2+1}(2x)\right]\]

Answer 40

can this be multiplied

Answer 41

if so how would it be done

Answer 42

No we wouldn't want to multiply that out :)
We would maybe want to simplify the part inside of the brackets though.
So it's written as a single fraction.

Answer 43

Er actually, if you leave it how it is, it SHOWS that you did logarithmic differentation.
Because we can clearly see the derivative of each logarithm in this form.

You probably don't want to simply too much further.

Answer 44

This is such a longggg process when you do logarithmic differentiation.
I hope I didn't make any silly mistakes in there :\

Answer 45

lemme plug it in and check for any errors

Answer 46

they are saying that it is incorrect

Answer 47

k one sec :)

Answer 48

thanks

Answer 49

I'm trying to check the problem on Wolfram, but they didn't use logarithmic differentiation. So the end result looks a lot different lol.

Answer 50

looking at answers similar to the problem yours seems correct. question is it xe^x^2 or is it xe^x2

Answer 51

You'd probably want to format it like this,
xe^(x^2)

Answer 52

If we can't get it to match up for some reason, let's just put in the form that results from doing the problem the other way.

I have the answer up on wolfram.

Answer 53

its alright thanks for the help. i got it. i missed formatted it. i really appreciate it

Answer 54

Ah ok good :) no prob