Find the derivative:
Hooray!! I have found it.. :P
\[\sqrt[4]{9-x^2}\]
Hahaha! Wait!
Thank God you're here. lol
You need to apply Chain Rule here.. :)
you must know this: \[\huge \sqrt[n]{a} = (a)^\frac{1}{n}\]
Step by step please. Haha
Yes, I'm familiar. :)
so, (9-x^2)^1/4 right?
So, please write your expression in this form first.. :)
I'm stuck there. Haha
\[(9-x^2)^{\frac{ 1 }{ 4}}\]
You must now know that : \[\large \frac{d}{dx}(f(x)^m) = m \cdot (f(x)^{m-1}). f'(x)\]
Just try to implement it here...
oh yeeeah... wait, is it now:\[\frac{ 1 }{ 4}(9-x^2)^{\frac{ -3 }{ 4}}\]
Have you forgot anything?? Yes, upto here you are right but it is still incomplete.. :P
How 'bout this? : \[\frac{ 1 }{ 4}(9-x^2)^{\frac{ -3 }{ 4}}(-2x)\]
You are getting it right.. :P
Good... :)
Is that correct? Haha, so what now? I just simplify?
Yes you can simplify it..
2 and 4 you can simplify.. You can again use : \[\large (a)^{\frac{-1}{n}} = \frac{1}{\sqrt[n]{a}}\]
I got it! Haha. Thaaanks again!
Hey, I have a question. Is there a name for that rule? The one I best response-d.
Which rule??
Oh, sh... I thought I'd only be best-ing one response. lol. The one with where you have to multiply it by f'(x)
Chain rule I said already... :)
I only know the part to bring down the coefficient. Oh, ok ok thanks. I'll look into the chain rule again, must've missed it in the lectures. Thanks!
Actually, from very first problem you are using that rule but you did not know that you were using that rule, Can I show you that??
Ok, sure. I know that there's a chain rule, but I thought it was just to cancel this and that.
Okay tell me what is the derivative of \(e^x\) with respect to x ??
\[xe^\] ?
\[xe^{x-1}\]
You have told this to me, I tolerated it, don't go outside and tell everybody that derivative of \(e^x\) is what you have just told, people will make fun of you.. :P
So, as you are just studying these things so I want to tell you that: \[\frac{d}{dx} (e^x) = e^x\]
If I have to tell you about these things, it will take time surely..
Oh, oh sorry. Wait, I'll just study first my lectures. Thanks again for your help!
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