anonymous
  • anonymous
Derived \[\sqrt[4]{x}\cos x\]
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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theEric
  • theEric
Hi! I'm not sure if I'll be able to help or not, yet. What is the question here? Do you know what the goal is?
anonymous
  • anonymous
no
raffle_snaffle
  • raffle_snaffle
Are we taking the derivative?

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theEric
  • theEric
Well, you say "Derived." When you derive something, you get to a new equation from some other equations. Oh, yeah! Do you mean we need to derivate that?
raffle_snaffle
  • raffle_snaffle
you mean differentiate
abb0t
  • abb0t
How did you derive that function?
anonymous
  • anonymous
yes dertivate @raffle_snaffle
raffle_snaffle
  • raffle_snaffle
|dw:1377970781565:dw|
raffle_snaffle
  • raffle_snaffle
so |dw:1377970825557:dw| apply product rule first.
abb0t
  • abb0t
oh, @raffle_snaffle you're such a smarty pants.
raffle_snaffle
  • raffle_snaffle
abbot you are the smarty pants, rated at a 94=P
raffle_snaffle
  • raffle_snaffle
Lol
abb0t
  • abb0t
1 Attachment
raffle_snaffle
  • raffle_snaffle
^ LOL abbot
theEric
  • theEric
So what raffle_snaffle did was define that \(\sqrt[4]{x}\cos x\) as the function of \(x\) that we call \(y(x)\), or just \(y\) for short. What you want is\[\frac{d\left(\sqrt[4]{x}\cos x\right)}{dx}\]which means you are finding the derivitive \(\sf with~respect~to~\it x\).
theEric
  • theEric
Do you have any idea where to start, to differentiate* (I used a bad word before) \(\sqrt[4]{x}\cos x\)?
theEric
  • theEric
@Muskan , it would help you to work with us! :)
raffle_snaffle
  • raffle_snaffle
muskan do you know how to apply product rule? lets say f(x)=(x)^1/4 and g(x)=cos(x) so product rule is f'(x)(g(x))+(g'(x)(f(x))
raffle_snaffle
  • raffle_snaffle
As you are apply product rule you must also use chain rule. I recommend you use limbnotes notation so you don't confuse yourself.
raffle_snaffle
  • raffle_snaffle
I butchered the guys name. Lol
theEric
  • theEric
Right. Your function, \(y(x)\) can be written as \(\sqrt[4]{x}\quad\times\quad\cos(x)\). both \(\sqrt[4]{x}\) and \(\cos(x)\) are each functions of \(x\). So raffle_snaffle named them \(f\) and \(g\) respectively. And since \(y=f\times g\), you can use the product rule to differentiate. Haha! I think it's Leibnitz, but not fully sure!
theEric
  • theEric
Leibniz*
theEric
  • theEric
(Quick search)
raffle_snaffle
  • raffle_snaffle
Ah yes that is how you spell it! Haha
theEric
  • theEric
And, @Muskan , raffle_snaffle is telling you to use chain rule, and you should! See, you have a function of a function of \(x\). \(g(x)=\cos(x)\).
theEric
  • theEric
I mean, chain rule applies there. It really applies everywhere. But here just know that \(\cos(\_\_\_)\) is a function on it's own. So, let's say that what is in there is \(h(x)\), which is another function of \(x\)! Then \(g(x)=\cos(h(x))\). Getting complicated, right? It might help to just look at it for a little bit! Then the chain rule is telling you that \[\frac{dg}{dx}=\left(\frac{dg}{dh}\right)\left(\frac{dh}{dx}\right)\] We see that \(\dfrac{dg}{dx}\) becomes \(\left(-\sin(h(x))\right)\left(\dfrac{dh}{dx}\right)\\=\left(-\sin(x)\right)\left(1x^0\right)\\=-sin(x)\) Do you know what \(\dfrac{df}{dx}\) might be?
theEric
  • theEric
Good luck!

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