My Math Forum Questions in derivatives

 Calculus Calculus Math Forum

 November 3rd, 2010, 07:39 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Questions in derivatives Hi all my answer a ) = x^2cosx + 2xsinx -2xsinx 2xcosx b) 4sinx/(cosx)^2 + -sex2x/(tanx)^2 c) sec2x/(1=tanx)^2 --------------------------------- part 2 a ) Qcosx+sinQ+cosQ b)xtan-2x^-1+7x c) one derivativ = -kmq^-2+h/2 second derivativ = 2kmq^-2
 November 3rd, 2010, 11:14 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Questions in derivatives Part 1 1. a) $y=x^2\cdot\sin x+2x\cdot\cos x-2\cdot\sin x$ $y'=x^2\cdot\cos x+2x\cdot\sin x-2x\cdot\sin x+2\cdot\cos x-2\cdot\cos x=x^2\cdot\cos x$ b) $y=\frac{4}{\cos x}+\frac{1}{\tan x}=4\cdot\sec x+ \cot x$ $y'=4\cdot\sec x\tan x-\csc^2x$ c) $p=\frac{\tan q}{1+\tan q}$ $p'=\frac{(1+\tan q)\cdot\sec^2q-\tan q\cdot(\sec^2q)}{(1+\tan q)^2}=\frac{\sec^2q}{(1+\tan q)^2}=\left(\frac{1}{\cos u+\sin u}\right)^2=\frac{1}{1+\sin(2q)}$ 2. $y=\csc x$ $y'=-\csc x\cot x$ $y''=-((\csc x)(-\csc^2x)+(\cot x)(-\csc x\cot x))=\csc^3x(1+\cos^2 x)$ Part 2 1. $r=\theta\cdot\sin\theta+\cos\theta$ $\frac{dr}{d\theta}=\theta\cdot\cos\theta+\sin\thet a-\sin\theta=\theta\cdot\cos\theta$ 2. $h(x)=x\cdot\tan(2\sqrt{x})+7x$ $h'(x)=x(\sec^2(2\sqrt{x})\frac{1}{\sqrt{x}})+\ tan(2\sqrt{x})+7=\sqrt{x}\sec^2(2\sqrt{x})+\tan(2\ sqrt{x})+7$ 3. $A(q)=\frac{km}{q}+cm+\frac{hq}{2}=\frac{h}{2}q+cm+ kmq^{-1}$ $\frac{dA}{dq}=\frac{h}{2}-kmq^{-2}$ $\frac{d^2A}{dq^2}=2kmq^{-3}$
 November 3rd, 2010, 09:25 PM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: Questions in derivatives Thanks

 Tags derivatives, questions

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post johngalt47 Calculus 2 November 23rd, 2013 11:44 AM jkh1919 Calculus 9 October 6th, 2012 03:21 PM r-soy Calculus 3 November 28th, 2010 11:31 AM r-soy Calculus 3 November 4th, 2010 02:42 PM r-soy Calculus 2 November 3rd, 2010 10:31 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top