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# derivative e 2x

Question:This is confusing with all the product rules, can't figure it out.

Answers:g(x) = x[e^(-2x)] Best to simplify this first. g(x) = x / e^(2x) Now you can solve for the first derivative. g'(x) = [e^(2x) * 1 - x * e^(2x) * 2] / [e^(2x)]^2 And simplify. g'(x) = [e^(2x) - 2x * e^(2x)] / [e^(2x)]^2 Now you can go for the second derivative. g''(x) = {[e^(2x)]^2 * 2e^(2x) - 2x * e^(2x) * 2 + e^(2x) * -2 - [e^(2x) - 2x * e^(2x)] * 2[e^(2x)] * e^(2x) * 2} / [e^(2x)]^4 Yes, it looks like a nightmare. I know. But you can simplify it. g''(x) = {2[e^(2x)]^2 - 4x - 2 - 2x * 2[e^(2x)] * 2} / [e^(2x)]^3 And this can still be simplified a bit more. g''(x) = {2[e^(2x)]^2 - 4x - 2 - 8x * e^(2x)} / [e^(2x)]^3 Okay, third derivative time. g'''(x) = {[e^(2x)]^3 * 4[e^(2x)] * e^(2x) * 2 - 4 - 8x * e^(2x) * 2 + e^(2x) * 8 - 2[e^(2x)]^2 - 4x - 2 - 8x * e^(2x) * 3[e^(2x)]^2 * e^(2x) * 2} / [e^(2x)]^5 Now you need to simplify this, and solve for a fourth derivative. I'll let you do that. I'll admit, it is confusing, but it gets less so when you simplify.

Question:Q1) (e^2x + 2x) ^0.5 Q2) (e ^ (x^0.5))+((e^x)^0.5) ans: (e^2x +1) / (e^2x +2x)^0.5 ans: (((e^x)^0.5)/2)+e ^(x^0.5)/2 (x^0.5) step by step working would really help - just learnt this stuff today. The problem with question 1 and 2 is that for question 1 my answer (which is wrong) was 1 / (e^2x + 1) ^0.5 and question 2 was (x^0.5)(e^(x^0.5)) + (1/(2e^x)) perhaps telling me whats wrong would be more helpful and how i should be doing it

Answers:In question one you have started correctly but not applied the chain rule (composition rule). After reducing the power of the bracket by one from a half to minus a half you should have put a half in front of the bracket and then multiplied by the derivative of the expression inside the bracket. So start with: (e^2x + 2x)^0.5 So reduce 0.5 to -0.5 becomes: 0.5 (e^2x + 2x)^-0.5 then times this by the derivative of the bracket. =0.5 (e^2x + 2x)^-0.5 * (2e^2x + 2) Note: The (2e^2x + x) is the derivative of (e^2x + 2x) So this simplifies to (e^2x + 1) / (e^2x + 2x)^0.5 Similarly for question 2. Start with: (e ^ (x^0.5)) + ((e^x)^0.5) First I'd rearrange the second part: (e ^ (x^0.5)) + (e^0.5x) Now lets take the derivative of the first part: = e^(x^0.5) * derivative of x^0.5 = e^(x^0.5) * 0.5 x^-0.5 Second bit is easy to find the derivative: e^(0.5x) * 0.5 So combining these two gives: e^(x^0.5) * 0.5 x^-0.5 + e^(0.5x) * 0.5 Neatening up a bit will give you the answer above. Note they have listed the two terms in the other order.

Question:Use the General Power Rule to find the derivative. y = e^(11-x+x^2)

Answers:The basic form for differentiating y = e^u is just y' = (e^u) * u'. In this case: y' = e^(11 - x + x ) * (-1 + 2x) y' = (2x - 1)e^(11 - x + x )

Question:For this purpose, the inflection point of the second derivative of f(x) must be 0. The problem I'm having is that I can't figure out a value for x between 0 and 2pi that will make f"(x)=0. In case the equation is not clear enough, f(x) equals to (e to the sine of x ) times (cos squared x minus sin x). Any help is immensely appreciated.

Answers:e to the power of anything cannot be 0, so you can just disregard that part. Solve for cos x - sinx = 0 instead. [Because sin x + cos x = 1] (1 - sin x) - sinx = 0 sin x + sinx - 1 = 0 Let sinx = a a + a - 1 = 0 Put that into the quadratic formula and you get: a = (-1 5)/2 Therefore sinx = (-1 5)/2 and just solve for x. Yeah, it's really complicated. I'm not sure if you differentiated it correctly. :S