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Use the definition of a derivative to find f (x) and f (x) 6 f(x) = Xf'(x)f"(x)...

Question

Use the definition of a derivative to find f (x) and f (x) 6 f(x) = Xf'(x)f"(x)

Use the definition of a derivative to find f (x) and f (x) 6 f(x) = X f'(x) f"(x)



Answers

Find the derivative of the function. $$f(x)=\frac{x}{e^{6 x}}$$

They give us the function after Lex is equal to X over E to the six X Well, let's write this a little. It's actually just x e to the negatives execs. So let's do product are derivative is equal Teoh. But I take the derivative of the first factor, not just the one. And then we leave the second factor alone. Then do the opposite this time Leave the first factor alone. Take the derivative of the second factor. So we'll have a negative 600. And if we factor out, are you to the negative six x We see that our function just either. The negatives XX times one minus six I was

In the human question we have to the potential defects. It was too well upon that last six. So to find the derivative, the formalized Lawen attached to zero therefore of explicit minus F. Of X divided by it. So he had I am providing the values the values are one of all Explosives Plus 6 -1 of our necks. Plus six. If you take calcium. Uh huh. The calcium should be explicit plus six, multiplied by X plus six. After that you get explicit minus of explosives plus six. So your X-plus six was cancel out. And finally you get yeah. Inaugural stands to zero one of college minus. Search upon Explicit less six multiple attacks. Let's six. After a translation of edge we put here limit. And after putting the limit we got minus one upon X plus X. X plus six. Holy square. So this is the final answer.

In this problem we want to find the second derivative of a given function. Why is equal to one divided by the square X minus six. This question is Charles you understanding of differentiation in particular how defy the derivative of a compound function? Like why? So to solve but we need to do is make use of the chain rule where DDX half of you of X equals D F D U d u D X. The chain rule compounds with the other shortcuts we learned. So if we were right, why is x minus six negative first power? We can easily use the chain rule where U. Is x minus six to you. Is rather F. Of U. Is U. To the negative. First from this will change the Y. The X equals DfB. You is negative X minus 65. 2nd time GTX one equals negative x minus six negative second. That's using chain. Well again we have the square Y. The x squared equals negative times negative two x minus six. Negative third times one. Which simplifies on the right as two over x minus six. Cute

For this problem, we are asked to use the limit definition of the derivative to find the derivative of F of X equals X to the power of 1/3. So to begin, we know that the limit definition specifically, it's the alternative limit definition that we want. So it's the limit as X approaches A. Of F. A. Or rather be fx is going to equal F of x minus F of a, divided by x minus A. So we have F of X equals X power one third. So we want to take the limit As X approaches a. of extra power of 1/3 -A. to the power of 1/3 divided by x minus a. Now we are given an additional note in the set up for the problem that we can rewrite the denominator here as being X to the power one moment here. So sorry about that, we can rewrite the denominator here that X minus a. As X to the power of 1/3 cubed minus a. To the power of 1/3 cubed. Now the reason why that's useful to us is we can now apply the difference of cubes in the denominator. So we take the limit as X approaches A of X power of one third minus eight to the power of one third. And then we can rewrite that denominator denominator excuse me? As X power of one third minus A. To the power of one third times. Uh x to the power of to over three plus uh X. It's the power of 1/3 A. To the power of 1/3 Plus A. to the power of to over three. Now the significance here we might be able to see is that there is a common factor of X power of 1/3 minus eight to the power 1/3 in numerator and denominator. So canceling that out or dividing that out, I should say gives us the limit as X approaches A. Of just X. Power to over three plus X. The power of one third. Eight to the power of one third Plus A. to the power of to over three. Now we can take this limit directly. We would get 1/8. The power of to over three plus. Now the X to the power of one third times eight. The power of one third becomes A. To the power of to over three and then we add on another A. To the power of to over three. So we are going to get 1/3 times a. To the power of to over three as the derivative.


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