Find a and b such that

- anonymous

Find a and b such that

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- anonymous

the following function is differentiable for all values of x.

- anonymous

|dw:1350070251929:dw|

- anonymous

I figured out a= 1 but I can't figure out what b is equal to.

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## More answers

- myininaya

a=1 ?

- anonymous

Oops. I figured out b = 1.

- anonymous

I can`t solve for a.

- myininaya

Yep. Now that is right so far.
You just set the limx->0 for both of them and solve for a b.
Good job.

- myininaya

Because one this it must need to be is continuous

- anonymous

Yep . I set the both as lim x-> 0 and I got that b = 1. I can't solve for a however.

- anonymous

Should I derive both functions and solve for a?

- anonymous

But that just means it's continuous. Not differentiable.

- myininaya

Right that is one condition we must need though

- myininaya

We need it to run smoothly when they meet (no sharp corners)...
hmm...
thinking....

- anonymous

So I go:
|dw:1350070954216:dw|

- anonymous

right? O_o .

- anonymous

No. Thats not right.

- anonymous

Never mind.

- myininaya

Ok...how about this :
What do you think of both parts of the function having the same exact slope at x=0
This will give us that smooth transition at x=0 that i'm talking about

- anonymous

Yep. I knew that. But unless I know what a is equal to I can't solve for the derivative at x=0.

- myininaya

\[(\arctan(ax+1))'|_{x=0}=(\frac{\pi}{4} e^{\sin(x)})'|_{x=0}\]

- anonymous

Right...

- anonymous

I tried that.

- myininaya

Find derivative of both sides
Set x=0
And solve
you are done

- anonymous

Let me try again. Maybe I made a mistake.

- anonymous

The derivative of arctan(ax+1)
|dw:1350071299130:dw|
right?

- myininaya

Is that an a on top ?

- anonymous

It shoud say 1/(1+(ax+1)^2)

- myininaya

no it should say a on top

- myininaya

that is your issue

- myininaya

(ax+1)'=a

- anonymous

Ohh. Why?

- anonymous

Ohh....

- anonymous

haha... Chain rule.... oops.

- myininaya

\[(\arctan(u(x))'=\frac{u'(x)}{{1+(u(x))^2}}\]

- anonymous

so it would be \[\frac{ a }{ 1+(ax+1)^2 }\]

- myininaya

Yep and the other side you have?

- anonymous

|dw:1350071580411:dw|

- anonymous

right?

- anonymous

I really appreciate the help by the way. Thanks a lot.

- myininaya

I think you got it from here :)
Great job.

- anonymous

I got a = pi/2

- anonymous

@myininaya .

- anonymous

Thanks for making this so easy to understand! :) .

- myininaya

That is what I got :)

- myininaya

Gj again.

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