Wednesday

April 16, 2014

April 16, 2014

Number of results: 1,447

**Anti-Derivatives**

find the antiderivative of f(x) = x^3(x-2)^2 Someone help me please!
*Thursday, April 26, 2012 at 3:32pm by I Need Help*

**Anti-Derivatives**

rather than resorting to some formula method, I would expand the whole thing, and then integrate each of the simple terms.
*Thursday, April 26, 2012 at 3:32pm by Reiny*

**Anti-Derivatives**

Would I square (x-2) first and then multiply it by x cubed, or x cubed multiplied by (x-2) and then square the answer? Order of operations is parenthesis, exponents, multiplication and division (left to right), and then addition and subtraction ( left to right), isn't it?
*Thursday, April 26, 2012 at 3:32pm by I Need Help*

**Anti-Derivatives**

you would square x-2 first, the squaring does not include the x^3, or else the whole thing would have been in brackets, should have had... x^3(x^2 - 4x+4) = x^5 - 4x^4 + 4x^2 now integrate that
*Thursday, April 26, 2012 at 3:32pm by Reiny*

**Anti-Derivatives**

Integrate? Meaning?
*Thursday, April 26, 2012 at 3:32pm by I Need Help*

**Calculus - Anti-Derivatives**

How would you find the Integral of (cos(x\8))^3, defined from -(4 x pi)/3) to (4 x pi)/3)
*Friday, March 14, 2008 at 5:55pm by Sean*

**Calculus - Anti-Derivatives**

cos^3(x)dx = cos^2(x)cos(x)dx = [1-sin^2(x)]cos(x)dx = [1-sin^2(x)]d(sin(x))
*Friday, March 14, 2008 at 5:55pm by Anonymous*

**Calculus - Anti-Derivatives**

Is cos(x)8 supposed to be cos 8x, (cosx)^8 or 8 cosx?
*Friday, March 14, 2008 at 5:55pm by drwls*

**Calculus - Anti-Derivatives**

Its actually supposed to be (cos(x/8))^3. With what Anonymous answered, I see he left out the x/8, instead only using x. Would you, Drwls, mind using x/8?
*Friday, March 14, 2008 at 5:55pm by Sean*

**Calculus - Anti-Derivatives**

Sorry, I missed the \ mark. It is always better to use / for fractions when typing. Without messing with trig identities for cos (x/8), let's just substitute u for x/8. Then your integral becomes Integral of (cos u)^3 = (1- sin^2 u) cos u du , from u = -(pi/6) to (pi/6) Now ...
*Friday, March 14, 2008 at 5:55pm by drwls*

**Calculus - Anti-Derivatives**

I can follow everything except when you change the (4 x pi)/3 to pi/6, and change pi/6 into 1/2. Could you clear that up for me?
*Friday, March 14, 2008 at 5:55pm by Sean*

**Calculus - Anti-Derivatives**

the upper limit for example was x = 4 pi/3 but u = x/8 so the upper limit using u instead of x is u = (4 pi/3) /8 or u = pi/6 then it changes again to go from u to v v = sin u so at that same upper limit where x = 4 pi/3 and u = pi/6, we need to find v so v at this upper limit...
*Friday, March 14, 2008 at 5:55pm by Damon*

**Calculus - Anti-Derivatives**

Oh. Thank you, Damon, and Drwls. That really helped me out.
*Friday, March 14, 2008 at 5:55pm by Sean*

**Derivatives**

What is f ''(x) and f ''''(x) of f(x)= e^-x^2 f"(x) = -e^-x^2(2x) F""(x)= -2e^-x^2 + 2xe^x^2 and also what is integral [0 to 2] f(x) dx
*Sunday, October 22, 2006 at 11:39am by 413*

**Derivatives**

What is the fourth derivative of f(x)= e^-x^2. I got f''(x)=-2e^-x^2 + 4x^2e^-x^2 i need some help on find f''''(x) You have f(x) = e-x2 then f'(x) = -2x*e-x2 = -2x*f(x) f"(x) = (4x2 - 2))e-x2 = f(x)*(4x2 - 2)) f"'(x)= f'(x)*(4x2 - 2)) + f(x)(8x) = -2x*f(x)*(4x2 - 2)) = f(x)(-...
*Tuesday, October 24, 2006 at 1:28pm by 413*

**derivatives**

Applications of derivatives A rectangle has its base on the x axis and its upper two vertices on the parabola y=12 - x^2. What is the largest area the rectangle can have, and what are its dimensions. Support your answer graphically. Thanks. Well, the parabola is symettric ...
*Saturday, November 18, 2006 at 2:16pm by Jen*

**derivatives**

find the derivative of y=2cotx+sec3x+cscX find d^2/dx^2 for 1+y=x+xy
*Wednesday, October 31, 2007 at 10:34pm by sara*

**derivatives**

The derivatives of these trig functions should be right there in your Calculus text. for the second of your questions, I will use y' for the first derivative and y" for the second derivative i.e. y'= dy/dx 1+y=x+xy so y' = 1 + y + xy' y'(1-x) = 1+y y' = (1+y)/(1-x) y" = [(1-x)...
*Wednesday, October 31, 2007 at 10:34pm by Reiny*

**Derivatives**

Find dy/dx for the function, xy-x-12y-3=0. Is the answer -14,I got this from 1-1-12=0, which solves to -14.
*Wednesday, November 5, 2008 at 12:33pm by George*

**Derivatives**

ydx+xdy-dx-12dy=0 y + x y' -12 y'=1 y'= (1-y)/(x-12)
*Wednesday, November 5, 2008 at 12:33pm by bobpursley*

**derivatives**

If the derivative of f is given by f'(x)=ex-3x2, at which of the following values of x does f have a relative maximum value?
*Saturday, January 1, 2011 at 11:00pm by lisa*

**derivatives**

set the derivative equal to zero and solve for x ex - 3x^2 = 0 x(e - 3x) = 0 x = 0 or x = e/3 use the 2nd derivative test to see which produces the maximum f ''(x) = e - 6x f "(0) = e f "(e/3) = e - 2e which is negative so the value of x = e/3 produces a relative maximum value...
*Saturday, January 1, 2011 at 11:00pm by Reiny*

**derivatives**

Explain how a long stock and long put strategy equals the cash flow from a long call strategy.
*Sunday, February 13, 2011 at 9:54pm by Josh*

**Derivatives**

Find the derivative of g(t)=8t^2+9t at t=1 algebraically... g'(1)= _________
*Monday, February 14, 2011 at 3:06pm by Cad*

**Derivatives**

If you are studying Calculus, this is about as simple and basic as it gets. You have to know this What have you got so far?
*Monday, February 14, 2011 at 3:06pm by Reiny*

**derivatives**

Compute f '(a) algebraically for the given value of a. HINT [See Example 1.] f(x) = −2x + 2; a = −5 f '(a) = 1Your answer is incorrect.
*Thursday, March 3, 2011 at 10:30pm by la*

**Derivatives**

a pationt recives an injection of 1.5 milligrams of a medication, and the amount remaining in the bloodstream t hours later is A(t)= 1.5e^-0.08t .find the instantaneous rate of change of this amount: a) Immediately after the injection (time t=0)
*Tuesday, June 28, 2011 at 12:16pm by Courtney *

**derivatives**

how do u find the derivative of (csc x)^7 -cos 4x???
*Thursday, October 27, 2011 at 11:57pm by hannah*

**derivatives**

You should have a list handy for the basic trig function derivatives so if y = (csc x)^7 -cos 4x dy/dx = 7(csc x)^6 ( -csc x)(cotx) + 4sin(4x) etc
*Thursday, October 27, 2011 at 11:57pm by Reiny*

**derivatives**

Thanks. My teacher told us we need to use the chain rule in these problems, and i'm having trouble figuring out when and where to use it.
*Thursday, October 27, 2011 at 11:57pm by hannah*

**derivatives**

I did use the chain rule in finding the derivative of (csc x)^7
*Thursday, October 27, 2011 at 11:57pm by Reiny*

**derivatives**

I know. I just realized that I messed up what goes where when I was doing it myself. My problem is applying it to other problems, esp when i also have to use the quotient or product rule.
*Thursday, October 27, 2011 at 11:57pm by hannah*

**derivatives**

I know. I just realized that I messed up what goes where when I was doing it myself. My problem is applying it to other problems, esp when i also have to use the quotient or product rule.
*Thursday, October 27, 2011 at 11:57pm by hannah*

**derivatives**

How do you find the derivative of (6 csc x)/x?
*Friday, October 28, 2011 at 1:29am by hannah*

**derivatives**

[ low*d(high) - high*d(low) ] / (low)^2 where high = numerator low = denominator d(high) and d(low) = respective derivatives first recall that the derivative of csc x = -(cot x)(csc x) therefore, [ x*(-6 (cot x)(csc x)) - 6 (csc x) ]/x^2 or -(6 csc x)*(x(cot x) + 1)/x^2 hope ...
*Friday, October 28, 2011 at 1:29am by Jai*

**derivatives**

How do I find the 2nd derivative of x^(-1/15)
*Friday, November 4, 2011 at 1:56am by Jacob*

**derivatives**

d/dx x^(-1/15) = (-1/15)x^(-16/15) (-1/15)d/dx x^(-16/15) = (16/225)x^-(31/15)
*Friday, November 4, 2011 at 1:56am by Damon*

**Derivatives**

differentiate and simplify y= (e^x - 1) / x^2
*Sunday, March 25, 2012 at 4:37pm by help*

**Derivatives**

d(uv)/dx= u dv/dx + v (du/dx) let u= e^x-1 du=e^x v= (x^-2) dv= -2/x I suspect you can simplify the final answer.
*Sunday, March 25, 2012 at 4:37pm by bobpursley*

**Derivatives?**

Given that f(-0.5)=2 and f'(-0.5)=4, use a tangent line approximation to estimate f(0). The lesson is not very clear on what a tangent line approximaton is.
*Tuesday, June 5, 2012 at 6:21pm by JULIA*

**Derivatives?**

The tangent line approximation is: f(x) = f(a) + (x-a)*f'(a) It is the formula for a straight line that goes through x = a with the slope of f(x) at x=a. Let a = -0.5 and x = 0 f(0) = 2 + (0.5)*4 = 4
*Tuesday, June 5, 2012 at 6:21pm by drwls*

**Derivatives**

Ship A is 70 km west of ship B and is sailing south at the rate of 25 km/hr.ship B is sailing north at the rate of 45 km/hr.how fast is the distance between the two ships changing 2 hours later?
*Monday, January 14, 2013 at 1:31am by Dani mak *

**Derivatives**

Make a sketch of their current position, placing B at the origin and A on the negative x-axis, 70 units from the origin At a time of t hours, let the position of Ship A be A1 and the position of ship B be B1 Now AA1 = 25t km , and BB1 = 45t km Join A1B1 Extend A1A upwards to C...
*Monday, January 14, 2013 at 1:31am by Reiny*

**Derivatives **

what is derivative of sin^2 (3x-1)^2 ? dy/dx (sin^2 (x))= 2 sin x cos x dy/dx (3x-1)^2= 6(3x-1) chain rule? = 2 sin ((3x-1)^2) cos ((3x-1)^2) * 6(3x-1) Please help. Thank you
*Thursday, March 14, 2013 at 11:28am by Mark*

**Derivatives**

I'm not sure if this is correct. Can someone clarify thanks.
*Thursday, March 14, 2013 at 11:28am by Mark*

**Derivatives**

you are correct Or, you could go on and say sin (2(3x-1)^2) * 6(3x-1)
*Thursday, March 14, 2013 at 11:28am by Steve*

**Derivatives**

I got 12 sin (3x-1)^2 cos (3x-1)^2 * (3x-1) as my final answer. Is this in the correct simplified form ?
*Thursday, March 14, 2013 at 11:28am by Mark*

**Derivatives**

it is ok, but you can use 2sinu cosu = sin(2u) to make it what I had: 6(3x-1) sin(2(3x-1)^2) If you want to retain your original argument of sin and cos, then leave it as you had it.
*Thursday, March 14, 2013 at 11:28am by Steve*

**Derivatives**

I have taken that into consideration; thank you!
*Thursday, March 14, 2013 at 11:28am by Mark*

**Derivatives**

with respect to x f(cos x) if f'(x)= 1/x find (f(cos x))' i got tan x sec x this is wrong. can you explain thank you
*Wednesday, March 20, 2013 at 12:21am by Maryanne*

**Derivatives**

I have gotten so many different responses from mine on this question Can someone PLEASE write out the steps for me for this. Thanks f'(x)= 2x/(x^2+1) Find the second derivative. I get 4x^3 + x^2- 12x + 1
*Monday, May 6, 2013 at 11:42am by Alaina*

**Derivatives**

Oops I meant 4x^3 + x^2 - 12x + 1/ ((x^2+1)^3)
*Monday, May 6, 2013 at 11:42am by Alaina*

**Derivatives**

how can you possibly get a polynomial as the derivative of a rational function? apparently f(x) = ln(x^2+1) f'(x) = 2x/(x^2+1) f"(x) = 2(1-x^2)/(x^2+1)^2 since if f =u/v f' = (u'v-uv')/v^2 so, the derivative of f' is ((2)(x^2+1) - (2x)(2x))/(x^2+1)^2 = (2x^2+2 - 4x^2)/(x^2+1)^...
*Monday, May 6, 2013 at 11:42am by Steve*

**Derivatives**

Find f'(x): (abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3)) So I thought I should split it into two parts: A and B. A) (x^2)/((2x-3)^2) B) ((3x+2)^(1/3))/(2x-3) My plan was to differentiate each part and then multiply them together, since multiplying their denominators together gives...
*Friday, May 24, 2013 at 1:35am by Erika*

**Derivatives**

X^2(3x-2)^-2/3/(2x-3)^3 - 2x(x+3)(3x+2)^1/3/(2x-3)^4
*Friday, May 24, 2013 at 1:35am by Kuai *

**Derivatives**

Find the derivatives of: 1. H(x)= sin2xcos2x The answer given is 2cos4x. My question is, how in the world did they get that!? Shouldn't the answer at least contain the sin function, either negative or positive seeing as it's the derivative of cos? Also, which rules are ...
*Thursday, October 3, 2013 at 3:52am by Daisy*

**Derivatives**

H(x) = sin2x cos2x = 1/2 sin4x That help? Using the product rule, it's a bit more work, but you can get the same answer: dH/dx = 2cos2x cos2x - 2sin2x sin2x = 2(cos^2 2x - sin^2 2x) = 2cos4x Don't forget your trig just because you're taking calculus! #2 df/dx = (u'v - uv')/v^2...
*Thursday, October 3, 2013 at 3:52am by Steve*

**derivatives**

x^2cosx-2xsinx-2cosx
*Tuesday, March 18, 2014 at 11:39pm by jnky*

**derivatives**

I assume that you want the derivative, since you titles your post that way. if y = x^2cosx-2xsinx-2cosx dy/dx = (x^2)(-sinx) + 2x cosx - 2x(-cosx) - 2sinx + 2sinx = -x^2( sinx) + 4x( cosx) etc if needed
*Tuesday, March 18, 2014 at 11:39pm by Reiny*

**derivatives**

yup! sloppiness on my part
*Tuesday, March 18, 2014 at 11:39pm by Reiny*

**Derivatives**

G(t)=5200/(1+12e^(-.52t)) Find G'(t) I keeo getting (-32448e^(-.52t))/(1+12e^(-.52t))^2 Is this correct?
*Wednesday, April 2, 2014 at 5:38pm by Andy*

**Derivatives**

write it as: G(t) = 5200 (1+12e^(-.52t)^-1 G ' (t) = -5200(1+12e^(-.52t) )^-2 (-.52)(12)e^(-.52t) = + 32448 e^(-.52t) / (1+12e^(-.52t) )^2 I got a positive instead of your negative.
*Wednesday, April 2, 2014 at 5:38pm by Reiny*

**math**

sec cube dx
*Tuesday, March 6, 2012 at 5:10am by anti derivative *

**math**

sec cube dx
*Tuesday, March 6, 2012 at 5:10am by anti derivative *

**math**

sec3 dx
*Tuesday, March 6, 2012 at 5:10am by anti derivative *

**American history**

do you own homework its not that difficult to read your text book... seriously
*Tuesday, November 13, 2012 at 2:47pm by anti cheating*

**anti derivative**

please help me find anti derivative of -sinx * cos^2x am i using integration by parts?
*Wednesday, November 19, 2008 at 2:47pm by mary*

**anti derivative**

No. Use substitution. Let u = cos x. Then du = -sin x dx The integral becomes that of u^2 du, which is u^3/3 After substituting back, that becomes (cos^3 x)/3
*Wednesday, November 19, 2008 at 2:47pm by drwls*

**Anti allergic**

Your site is very convenient in navigation and has good design. Thanks!. I am from Guinea and too poorly know English, tell me right I wrote the following sentence: "Background tmk is being developed as an anti allergic drug having both lipoxygenase inhibitory activity and ...
*Thursday, July 20, 2006 at 7:06am by Von*

**anti derivative**

hi
*Wednesday, November 19, 2008 at 2:47pm by Anonymous*

**Anti-ln?!?!**

use a calculator to find the anti-ln(-0.049) to the nearest ten thousandth. My calculator doesn't do anti-ln. just ln. Someone helo me please?
*Saturday, April 21, 2012 at 1:47pm by Tabby*

**Anti-ln?!?!**

Use the inverse key INV LN (-.049) or do this use the e^ key anti-ln(a)= e^a
*Saturday, April 21, 2012 at 1:47pm by bobpursley*

**Anti-ln?!?!**

Is the answer 0.9521811297? That's what I got using the e^ key. (I don't have an inverse key on my calculator. It's a Casio fx-9750G PLUS Power Graphic, and the manual for it is literally one inch thick with 453 pages.)
*Saturday, April 21, 2012 at 1:47pm by Tabby*

**Anti-ln?!?!**

correct. Put this in your google search window: e^-.049=
*Saturday, April 21, 2012 at 1:47pm by bobpursley*

**Anti-ln?!?!**

Ah. Thank syou o much for your time and help. God Bless you.
*Saturday, April 21, 2012 at 1:47pm by Tabby*

**Anti-ln?!?!**

Tabby, I had a Casio at one time Look for a 2ndF key, it works the same way enter: 2ndF ln -.049 =
*Saturday, April 21, 2012 at 1:47pm by Reiny*

**Anti-ln?!?!**

Thanks Reiny!
*Saturday, April 21, 2012 at 1:47pm by Some1 Help Me Plz*

**Math grade 12 **

Derivatives
*Saturday, August 13, 2011 at 9:59am by Atul prakash patil*

**calculus anti-derivation**

Given that the graph of f(x) passes through the point (2,5) and that the slope of its tangent line at (x, f(x))is 3x+5, what is f(3) ?
*Saturday, April 14, 2012 at 11:15pm by lindsay*

**calculus anti-derivation**

if dy/dx = 3x+5 , then y = (3/2)x^2 + 5x + c when x=2, y = 5 5 = (3/2)(4) + 10 + c 5 = 6 + 10 + c c = -11 f(x) = (3/2)x^2 + 5x - 11 f(3) = (3/2)(9) +15 - 11 = 35/2
*Saturday, April 14, 2012 at 11:15pm by Reiny*

**anti-derivative help!!**

Given f"(x)=-8sin3x and f'(0)=-1 and f(0)=-6. Find f(pi/5). i got stuck with f(pi/f) f"(x)=-8sin3x f'(x)=8cos3x+c f'(0)=8cos3(0)+c=-1 c=-9 f'(x)=8cos3x-9 f(x)=-8sin3x-9x+c f(0)=-8sin3(0)-9(0)+c=-6 c=-6 f(pi/5)-8sin3(pi/5)-9(pi/5)-6. so am i right?
*Saturday, April 14, 2012 at 11:40pm by lindsay*

**anti-derivative help!!**

i mean i got stuck with f(pi/5)
*Saturday, April 14, 2012 at 11:40pm by lindsay*

**anti-derivative help!!**

no, if f' = sin(nx) f = -1/n cos(nx) remember the chain rule: d/dx cos u = -sinu du/dx so, ∫sin(nx) = -1/n cos(nx) So. Let's see what we have: f''(x) = -8sin3x f'(x) = 8/3 cos3x + c -1 = 8/3 + c c = -11/3 f'(x) = 8/3 cos3x - 11/3 f(x) = 8/9 sin3x - 11/3 x + c -6 = 0 - 0...
*Saturday, April 14, 2012 at 11:40pm by Steve*

**Math-Derivitives **

if f(x)=x^3+5x, then f'(2)= if f(x)=cos(2x), then f'(pi/4) derivatives and anti derivatives please help
*Tuesday, March 23, 2010 at 10:18pm by AdrianV*

**Calculus/Derivatives**

In addition, can you walk me through how to get the derivatives for these 2 statements, too? a) y = x^5/3 - 5x^2/3 b) y = (the cubed root of the quantity) [(x^2 - 1)^2] Hi there. I need to find the first derivative of this statement. y=x(x+2)^3 I tried the chain rule, but I ...
*Wednesday, March 14, 2007 at 9:07pm by Amy*

**calculus - derivatives**

can you please find the first 5 derivatives for: f(x) = (0.5e^x)-(0.5e^-x) f'(x) = ? f''(x) = ? f'''(x) = ? f''''(x) = ? f'''''(x) = ? thanks :) f(x) = (0.5e^x)-(0.5e^-x) f'(x) = 0.5 e^x + 0.5 e^-x f''(x) = 0.5 e^x - 0.5 e^-x f'''(x) = 0.5 e^x + 0.5 e^-x f''''(x) = 0.5 e^x - 0...
*Monday, July 30, 2007 at 6:15pm by COFFEE*

**Calculus: Derivatives**

I'm having trouble understanding what derivatives in calculus (instantaneous speed) is. Can someone please explain it to me, or provide some easy links to learn from? Thanks!
*Sunday, September 9, 2007 at 8:32pm by Joe*

**Calculus: Derivatives**

a derivative is a rate of change. As speed is the rate of change of distance. http://www.themathpage.com/aCalc/derivative.htm
*Sunday, September 9, 2007 at 8:32pm by bobpursley*

**Calculus - Derivatives**

I need help finding the derivative of 100P^-2 I know that 100P^2 would be 200P, but the -2 is throwing me off. Could someone help me out here? Thanks.
*Wednesday, October 3, 2007 at 11:39am by Adam*

**Calculus - Derivatives**

exactly the same multiply the the coeffecient by exponent and subtract one from the exponent. -200 P ^-3
*Wednesday, October 3, 2007 at 11:39am by Quidditch*

**Calculus - Derivatives**

Oh, okay. I got it. Thanks!
*Wednesday, October 3, 2007 at 11:39am by Adam*

**Calculus - Derivatives**

Ok, now I have to use the derivative and solve for P. So, the equation is: 4 = -200P^-3 I can get it down to: -50 = P^-3 But, I don't know how to get the -3 away from the P.
*Wednesday, October 3, 2007 at 11:39am by Adam*

**Calculus - Derivatives**

Check your work above. There is an error in the value on the left side. Fix that before the next step.
*Wednesday, October 3, 2007 at 11:39am by Quidditch*

**Calc-Derivatives**

I asked this before and did not get a response so I am trying again. (Note, I didn't know how to type the symbol for pi) What is the derivative of: 2pi^2r^4 would it simply be 8pi^2r^3? I was thinking the 2pi^2 is simply the equivalent of a constant in front of the r^4 and so ...
*Wednesday, November 28, 2007 at 8:38pm by Rob*

**Calc-Derivatives**

correct, your thinking is right on
*Wednesday, November 28, 2007 at 8:38pm by Reiny*

**Math, derivatives**

f(x) = x² + 2Cos²x, find f ' (x) a) 2(x+cos x) b) x - sin x c) 2x + sin x d)2(x - sin2x) I got neither of these answers, since the 2nd part should be chain rule, right? f(x) = x² + 2Cos²x = x² + 2(Cos x)² then f '(x) = (2)(2)(cosx)(-sinx) My answer: f '(x) = 2x - 4cosxsinx If ...
*Monday, May 5, 2008 at 8:19pm by Terry*

**Math, derivatives**

sorry, this part: then f '(x) = (2)(2)(cosx)(-sinx) should say: then f '(x) = 2x + (2)(2)(cosx)(-sinx)
*Monday, May 5, 2008 at 8:19pm by Terry*

**Math, derivatives**

You're on the right track. sin 2x= 2sin x cos x (identity) substitute into your first answer. f'(x)=2x-2(sin2x) f'(x)=2(x-sin2x)
*Monday, May 5, 2008 at 8:19pm by Doug*

**Math, derivatives**

Thanks for your reply! That makes much more sense now. I guess I should go and review my identities.
*Monday, May 5, 2008 at 8:19pm by Terry*

**Math, derivatives**

Let g(x) = sin (cos x^3) Find g ' (x): The choices are a) -3x^2sinx^3cos(cos x^3) b) -3x^2sinx^3sin(cos x^3) c) -3x^2cosx^3sin(cos x^3) d) 3x^2sin^2(cos x^3) I'm not exactly sure where I should start. Should I begin with d/dx of sin? Or do the inside derivative first...and do ...
*Sunday, May 11, 2008 at 6:27pm by Marissa*

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