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November 4th, 2010, 10:24 AM   #1
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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
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November 4th, 2010, 11:36 AM   #2
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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 double-angle 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., 999897...321, 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.
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November 4th, 2010, 03:19 PM   #3
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Re: Questions in derivatives

censored^2... lol
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November 4th, 2010, 03:42 PM   #4
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Re: Questions in derivatives

Now don't go off on a tangent...let's secant we do this!
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