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 First, we find y'(x): Now we use : Thus, the tangent line is horizontal. 2.) You use the power and chain rules: Recall the doubleangle identity for sine: 3.) edit: Sorry, I missed the minor mistake you made. 4.) I am assuming you mean First, let's use the identity: to rewrite y as 5.) edit: missed another minor mistake! 6.) At x = 4, we have 7.) Yes, since acceleration is constant (assuming no forces but gravity) then jerk would have zero magnitude. 8.) 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. Note that the domain is all reals. We can see that and does not exist. Thus, those values for x are the critical points for f. 10.) 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: Solve for y'. 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 . Now, by applying the definition of the secant function which is the hypotenuse divided by the adjacent side, we have 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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