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# Differentiating lnx watch

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1. hi, need some help with this question
find the equation of the normal of ln(x^2+1) where x = 1

for dy/dx, i got 2x/(x^2+1) = 0, and I got 1 for the gradient, -1 for the gradient of the normal, im trying to find the y coordinate but im getting a faction when I put x into the original equation, am i doing something wrong?

also find the coordinates of the point where the gradient of y = xlnx is 0,
i got 1 + lnx for dy/dx, = 0, but how do i find the x value?

thanks
2. (Original post by jacksonmeg)
hi, need some help with this question
find the equation of the normal of ln(x^2+1) where x = 1

for dy/dx, i got 2x/(x^2+1) = 0, and I got 1 for the gradient, -1 for the gradient of the normal, im trying to find the y coordinate but im getting a faction when I put x into the original equation, am i doing something wrong?

also find the coordinates of the point where the gradient of y = xlnx is 0,
i got 1 + lnx for dy/dx, = 0, but how do i find the x value?

thanks
You have a function

You are correct It is equal to dy/dx not 0

You are right -1 is the gradient of the normal so you now need to find the value of y when

When (it is not a fraction, it is a log you will end up with)
3. (Original post by Robbie242)
You have a function

You are correct It is equal to dy/dx not 0

You are right -1 is the gradient of the normal so you now need to find the value of y when

When (it is not a fraction, it is a log you will end up with)
ln(2) = decimal ... idk what im doing wrong
4. (Original post by jacksonmeg)
ln(2) = decimal ... idk what im doing wrong
That's right when that's it

You have a gradient now and a point (1,ln2)

Now you can find the equation of the normal by substituting this into a straight line equation
5. (Original post by jacksonmeg)
ln(2) = decimal ... idk what im doing wrong
With C3, you need to learn to keep things like ln as they are. ln2 is ln2, use the formula y - y1 = m(x - x1).
6. (Original post by Robbie242)
That's right when that's it

You have a gradient now and a point (1,ln2)

Now you can find the equation of the normal by substituting this into a straight line equation
alright I got y = -x+ln2+1 which i checked in the back and is correct
could you explain how to do the second one please?
7. (Original post by TSR561)
With C3, you need to learn to keep things like ln as they are. ln2 is ln2, use the formula y - y1 = m(x - x1).
I know getting a lot of e^x and lnx questions wrong by substituting values in and getting decimals ...
8. (Original post by jacksonmeg)
alright I got y = -x+ln2+1 which i checked in the back and is correct
could you explain how to do the second one please?
Sure you need to find the value of x and the corresponding value of y such that

Well we know that

We need to know when the gradient is zero so

Means

Now you need to remember that ln is the inverse of e and vice versa

So

then once you have x, plug it back into the equation for y which is
9. (Original post by jacksonmeg)
I know getting a lot of e^x and lnx questions wrong by substituting values in and getting decimals ...
You'll get used to it haha. It's especially fun when you're integrating and can't use the integrating bit of your calculator (if you have one) to just give you the answer
10. (Original post by Robbie242)
Sure you need to find the value of x and the corresponding value of y such that

Well we know that

We need to know when the gradient is zero so

Means

Now you need to remember that ln is the inverse of e and vice versa

So

then once you have x, plug it back into the equation for y which is
you've lost me at the e^ln bit :/
11. (Original post by TSR561)
You'll get used to it haha. It's especially fun when you're integrating and can't use the integrating bit of your calculator (if you have one) to just give you the answer
we started integrating sin, cos, tan, ln, e^x and area under a curve last week, I prefer this to differentiation so far >:/
12. (Original post by jacksonmeg)
you've lost me at the e^ln bit :/
I think he's basically saying ln(x) is the same as loge(x)... so if loge(x) = -1, x = ...
13. (Original post by jacksonmeg)
you've lost me at the e^ln bit :/
It's a general rule that and as they are inverses of each other

You have

Raise both sides to the power of e

This gives:

Spoiler:
Show

Which means:

14. (Original post by Robbie242)
It's a general rule that and as they are inverses of each other

You have

Raise both sides to the power of e

This gives:

Spoiler:
Show

Which means:

where has the ln gone? :l
and the e
15. (Original post by jacksonmeg)
where has the ln gone? :l
It's the inverse so both ln and e cancel each other leaving only the behind in this case
If you don't understand this you might need to read chapter 3 of your C3 book, it has explanations on this.
16. (Original post by Robbie242)
It's the inverse so both ln and e cancel each other leaving only the behind in this case
If you don't understand this you might need to read chapter 3 of your C3 book, it has explanations on this.
alright ill take a look at it, can't wait to finish maths, abandon the ship before it sinks
17. (Original post by Robbie242)
It's the inverse so both ln and e cancel each other leaving only the behind in this case
If you don't understand this you might need to read chapter 3 of your C3 book, it has explanations on this.
the answer is (1/e, -1/e), not sure how to get y, something to do with reflection in y=x? cant find explanation of it in the book
18. (Original post by jacksonmeg)
is it (1/e, -1/e) ?
because inverse graphs are reflections of each over in y=x, or is my reasoning completely wrong? xD
The x-coordinate is correct

Let me find the y-coordinate

when ,

Using log laws we have

so

I'm not sure about reasoning actually but working out the actual coordinate gives the answer the book has
19. (Original post by Robbie242)
The x-coordinate is correct

Let me find the y-coordinate

when ,

Using log laws we have

so

I'm not sure about reasoning actually but working out the actual coordinate, you are correct!
not correct! I read the back of the book, I dont understand logs at all
20. (Original post by jacksonmeg)
not correct! I read the back of the book, I dont understand logs at all
Then read chapter 3 and do questions from there on logs, there's not point jumping into differentiation if you don't know your logs

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