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 November 4th, 2010, 09:24 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Questions in derivatives Questions in derivatives hi all the slope of the tangent line to the curve y = pi sinc at x = pi/2 = ... I can't understand like this question now I do derivative y = cosx then what the next step .... ? 2 ) the derivative of y cos^2X = here did I use product rule? and what is the derivative of cos^2 ? 3 ) the derivative of y =cos3x == > -3sinx 4 ) derivative of sex^2(pi + 3t) did I bring pow to 2sex(pi + 3t ) 5 ) if the velocity of a particle at any time t is given by the equation v(t) = sint - cost then its acceleration ...... ===> -cost+sint 6 ) the slope of the tangent to the curve y=rootx at x =4 derivative = 2/2rootx 7 ) the jerk during free fall is ..... o m/s^3 8 ) if y = x^99 + 8 the d100(y)/dx100 = ............... 99 9 _ the critical point of the function f(x) = x^1/3(x+3) ..... I think no critical point 10 ) If x = sin y then dy/dx .... y= cosx
 November 4th, 2010, 10:36 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 1.) I am assuming you mean $y=\pi\cdot\sin x$ First, we find y'(x): $y'(x)=\pi\cdot\cos x$ Now we use $x=\frac{\pi}{2}$: $y'\left(\frac{\pi}{2}\right)=\pi\cdot\cos\left (\frac{\pi}{2}\right)=0$ Thus, the tangent line is horizontal. 2.) $y=\cos^2x=(\cos x)^2$ You use the power and chain rules: $y'=2\cdot\cos x(-\sin x)=-2\cdot\sin x\cos x=-\sin(2x)$ Recall the double-angle identity for sine: $\sin(2\theta)=2sin\theta\cos\theta$ 3.) edit: Sorry, I missed the minor mistake you made. $y=\cos(3x)$ $y'=-\sin(3x)\cdot(3)=-3\sin(3x)$ 4.) I am assuming you mean $y=\sec^2(\pi +3t)$ First, let's use the identity: $\sec(\theta+\pi)=-\sec(\theta)$ to rewrite y as $y=\sec^2(3t)$ $y'=2\cdot\sec(3t)(\sec(3t)\cdot\tan(3t))(3)=6\ cdot\sec^2(3t)\tan(3t)$ 5.) edit: missed another minor mistake! $v(t)=\sin t-\cos t$ $a(t)=v#39;(t)=\cos t+\sin t$ 6.) $y(x)=\sqrt{x}=x^{\frac{1}{2}}$ $y'(x)=\frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$ At x = 4, we have $y'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}$ 7.) Yes, since acceleration is constant (assuming no forces but gravity) then jerk would have zero magnitude. 8.) $y=x^{\small{99}}+8$ $\frac{d^{\tiny{100}}y}{dx^{\tiny{100}}}=0$ The 99th derivative would be 99! (that's 99 factorial, i.e., 99·98·97·...·3·2·1, a constant, albeit huge), but one more differentiation would still yield zero. 9.) A critical point of a function f is a number c in its domain for which f'(c) = 0 or f'(c) does not exist. $f(x)=x^{\tiny\frac{1}{3}}(x+3)$ Note that the domain is all reals. $f'(x)=x^{\tiny\frac{1}{3}}(1)+(x+3)\left(\frac {1}{3}\cdot x^{\tiny-{\frac{2}{3}}}\right)=\frac{4x+3}{3x^{\tiny{\frac{ 2}{3}}}}$ We can see that $f'\left(-\frac{3}{4}\right)=0$ and $f'(0)$ does not exist. Thus, those values for x are the critical points for f. 10.) $x=\sin y\:\therefore\:y=\sin^{\tiny{-1}}x$ $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$ You probably haven't been given the derivatives of the inverse trigonometric functions yet, so we can derive this result in the following way: Implicitly differentiate x with respect to x: $1=(\cos y)y#39;$ Solve for y'. $y'=\sec y=\sec(\sin^{\tiny{-1}}x)$ $\sin^{\tiny{-1}}x$ denotes an angle in a right triangle for which the side opposite is x and the hypotenuse is 1. By the Pythagorean theorem we find that the side adjacent is $\sqrt{1-x^2}$. Now, by applying the definition of the secant function which is the hypotenuse divided by the adjacent side, we have $y'=\frac{1}{\sqrt{1-x^2}}$ edit: Please look at problems 3 and 5, the first time I reviewed your work I did not catch the minor mistakes.
 November 4th, 2010, 02:19 PM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Questions in derivatives censored^2... lol
 November 4th, 2010, 02:42 PM #4 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 Now don't go off on a tangent...let's secant we do this!

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# y=99x^99 find d^100y/dx100

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