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derivative e 2x

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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

From Youtube

Derivatives of Inverse Trig Functions :Worked problem in calculus. Find the derivatives of (a) g(x) = xsin^{-1}(x^2+x), (b) h(x) = tan^{-1}(ln(x)), and (c) s(x) = sec^{-1}(e^2x).

Calculus: Derivatives of Exponential Functions :www.mindbites.com The power rule for differentiation does not apply to exponential functions (eg the derivative of 2^x does NOT equal x*2^(x-1)). In this lesson, we will return to the limit definition of the derivative to discover how to differentiate exponential functions like 2^x. Professor Burger will graph the exponential function and calculate the slope of the tangent line. In the end, we will arrive at the fact that the derivative of an exponential function is the product of the function and the natural log of its base. You will also see how to arrive at e = 2.71828..., which is the only constant for which its derivative, when raised to any power is equal to the same thing. Thus, the derivative of e^x is equal to e^x and the e^(ln x) = e. The derivative of N^x is equal to N^x*ln N. The constant 'e' turns out to be very important in math, and Dr. Burger will explain some of its uses as a critical constant. Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at www.thinkwell.com The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'H pital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP ...