+9673 I have 3 reasons that I do not use the 'Post the Question' function(called 'the function' below) properly. Moderators and Admins, please look at my explanation of not using the function and DO NOT take action(such as deleting this post or banning my IP from the forum) immediately.
1. The LaTeX does not work in the Private Message section.
2. I cannot use plain text to type Maths, it will confuse me.
3. I cannot find his original question once again scrolling down the forum.
dho0o0m's question:
\(1. f(x) = 5\pi^2,\;\text{find }f'(x)\\ 2.f(x) = 5\sqrt x,\text{ find }f'(x)\\ 3.f(x)=\dfrac{1}{x^4} ,\text{ find }f'(x)\\ 4. f(x) = 2x^7 - 5x^{-7} ,\text{ find }f'(x)\\ 5. f(x) = \dfrac{2x-1}{x^3} ,\text{ find }f'(1)\\ 6. f(x) = 3x^4\ln x ,\text{ find }f'(1)\\ 7.f(x) = 8\sqrt x + 6x^{3/4},\text{ find }f'(x)\\ 8.y = \dfrac{5}{x^2}-\dfrac{3}{\sqrt x}+5x^4,\text{ find }\dfrac{dy}{dx}\\ 9. y = (6x^3+2)(5x-3),\text{ find }\dfrac{dy}{dx}\\ 10. f(t)=\dfrac{(8t-3)(2t+5)}{t-7},\text{ find }f'(1)\)
My answer:
\(1.\\ f(x) = 5\pi^2\\ f'(x) = 0\text{ as it is a constant}\\ ......................................................\\ 2.\\ f(x) = 5\sqrt x\\ f'(x) = \dfrac{5}{2\sqrt x}\\......................................................\\ 3.\\ f(x) = \dfrac{1}{x^4}=x^{-4}\\ f'(x) = -4x^{-5}=-\dfrac{4}{x^5}\\......................................................\\ 4.\\ f(x) = 2x^7 - 5x^{-7}\\ f'(x) = 14x^6 +35x^{-6}=14x^6 + \dfrac{35}{x^6}\\......................................................\\ 5.\\ f(x) = \dfrac{2x-1}{x^3}=\left(x^{-3}\right)(2x-1)=2x^{-2}-x^{-3}\\ f'(x) = -4x^{-3} +3x^{-4}=-\dfrac{4}{x^3}+\dfrac{3}{x^4}\\ f'(1) = -\dfrac{4}{1^3}+\dfrac{3}{1^4}=-4 + 3 = -1 \\\text{P.S.: Need not use quotient rule nor product rule}\\ ......................................................\\ 6.\\ f(x) = 3x^4 \ln x\\ f'(x) = (12x^3)(\ln x)+(3x^4)(\dfrac{1}{x})=12x^3\ln x + 3x^3\\ f'(1) = 12(1^3)\ln 1 + 3(1^3)=12(0) + 3(1) = 3 \\......................................................\\ 7.\\ f(x)=8\sqrt x + 6x^{3/4}=8x^{1/2}+6x^{3/4}\\ f'(x) = \dfrac{4}{\sqrt x}+\dfrac{9}{2x^{1/4}}\\......................................................\)
\(8.\\ y=\dfrac{5}{x^2}-\dfrac{3}{\sqrt x}+5x^4 = 5x^{-2} - 3x^{-1/2} + 5x^4\\ \dfrac{dy}{dx} = -10x^{-3} +\dfrac{3}{2x^{3/2}}+20x^3 = -\dfrac{10}{x^3}+\dfrac{3}{2x^{3/2}}+20x^3\\......................................................\\ 9.\\ y = (6x^3 + 2)(5x - 3)\\ \dfrac{dy}{dx} = (18x^2)(5x - 3) + (6x^3 + 2)(5) = 120 x^3 -54x^2+10\\......................................................\\ 10.\\ f(t) = \dfrac{(8t-3)(2t+5)}{t-7}= \dfrac{16t^2 +34t-15}{t-7} = 16t + 146 +\dfrac{1007}{t-7}\\ f'(t) = 16 - 1007(t-7)^{-2}=16-\dfrac{1007}{(t-7)^2}\\ f'(1) = 16 - \dfrac{1007}{(1-7)^2}=16 - 27\dfrac{35}{36} = -\dfrac{431}{36}\)
I am really sorry for people who thought there was a new question...... I am sorry for disrupting the forum's questioners and answerers just for 1 person. I deeply apologize here. Sorry.
~The smartest cookie in the world
+37146 # 4 answer should be 14x^6 +35/(x^8) is the only issue I see....but I am not an expert.
Another method for 9:
(6x^3 +2) (5x-3) = (multiply it out)
30x^4 -18x^3 +10x -6 now d/dx =
f'(x) = 120 x^3 -54x^2 + 10
+37146 No problem or flak from me 'cookie'....I learn a lot from other people's answers ! Thanx
+118673 It is not a problem Max :)
Having Latex in the message section would be good but I don't see it as a major issue because maths should really be on the forum anyway. What you have done is perfectly fine :)
An easy way to message latex is to start a new question, write your latex, copy the result with a screen capture, gyazo is a great program for this (although it does play up sometimes) and then just post the clip as a picture. If all works properly it does not take much extra time.
Of course you would never finish posting the new message, it was just a tool. :)
BUT
Putting it on the forum like this was the best idea. :)
+37146 # 4 answer should be 14x^6 +35/(x^8) is the only issue I see....but I am not an expert.
Another method for 9:
(6x^3 +2) (5x-3) = (multiply it out)
30x^4 -18x^3 +10x -6 now d/dx =
f'(x) = 120 x^3 -54x^2 + 10
