How do I find the derivative of [math]\sin(2x) \cos^2 x[/math], btw the anwer is [math]-6\sin^2 x\cos^2 x + 2\cos^4 x?[/math]
Do you know what? I really don't like the product rule and I love trigonometry :) So let's not use product rule, sorry everyone else :) what you need to know is the two trigonometric formulas you should always know:[math] sin(2x)=2*sin(x)*cos(x) [/math] and [math] cos(2x)=2cos^2(x)-1 [/math]but I will use them in the following forms:[math] sin(4x)/2=sin(2x)*cos(2x) [/math] and [math] cos^2(x)=(cos(2x)+1)/2 [/math]Maybe you can where I am getting at :) instead of your expression, by using my second trigonometric formula, I can write:[math] sin(2x)(cos(2x)+1)/2 [/math] developping it a bit:[math] sin(2x)*cos(2x)/2 + sin(2x)/2[/math] now the first term is familiar. We will use the first trigonometric formula:[math] sin(4x)/4 + sin(2x)/2 [/math] wow it looks so simple now. Then we can differentiate and we find :[math] 4*cos(4x)/4 + 2*cos(2x)/2 = cos(4x) + cos(2x) [/math]I know it is not the same answer you are looking for but it actually is!
What is the derivative of cos^2 x/2?
Let y = cos^2 x/2Now differentiating w.r.t.xdy/dx = d/dx ( cos^2 x/2)Or, dy/dx = 2 cos(x/2).d/dx (cosx/2)Or, dy/dx = 2 cos(x/2).(-sinx/2).d/dx(x/2)Or, dy/dx = - 2 cos(x/2).sin(x/2).(1/2)Or, dy/dx = - sinx.(1/2) = - 1/2 sinx[Since sin2x= 2sinx.cosx]So the answer is -1/2 sinxThank you
What is the derivative of (cos x-sin x)/(cos x+sin x)?
Let y=(cosx-sinx)/(cosx +sinx)y=(1-tanx)/(1+tanx)y=tan(pi/4 - x)……..{·.· tan(pi/4) =1 }{note:tan(a-b) =[ tan(a)-tan(b)]/[1 + tan(a).tan(b) ]}Diff.w.r.t x.·. y' = - sec^2( pi/4 -x) ……{ derivative of (pi/4 -x) w.r.t x is (-1) }Your ans is -sec^2(pi/4 -x) .
AP Calc AB: Help finding where the derivative fails to exist?
Here's the question: [Let g(x) = f(x) * tan(x) + kx, where k is a real number. It is known that f is differentiable for all x, and f(pi/4) = 4, f'(pi/4) = 2. a) For what values of x, if any, in the interval 0 < x < 2pi will the derivative of g fail to exist? Justify. b) If g'(pi/4) = 6, find the value of k.] I mostly need help with part a. Thanks in advance!
CALCULUS consider the equation 3sinx+3cosx from 0<=x<=2pi?
I've been doing this problem over and over but I keep getting it wrong. all i know is that f increases (0,pi/4) and (5pi/4,2pi) and that it decreases (pi/4,5pi/4) a. find local minimum and maximum b. find inflection points c. find the intervals where f is concave up d. find intervals where f is concave down thank you so much in advance!
F(x) =8x sinx cosx f '(pi/2) =?
f(x) = 8x * sin(x) * cos(x) f(x) = 4x * 2 * sin(x)cos(x) f(x) = 4x * sin(2x) f'(x) = 4x * 2 * cos(2x) + 4 * sin(2x) f'(x) = 8x * cos(2x) + 4 * sin(2x) f'(pi/2) = 8 * (pi/2) * cos(2 * pi/2) + 4 * sin(2 * pi/2) f'(pi/2) = 4 * pi * cos(pi) + 4 * sin(pi) f'(pi/2) = 4 * pi * (-1) + 4 * 0 f'(pi/2) = -4pi
F(x)= cos^2 x - 2sinx, 0≤x≤2pi?
f(x)= cos^2 x - 2sinx = (cos x)^2 - 2sin x, 0≤x≤2pi f '(x) = 2cos x * -sin x -2cos x = 2cos x (-sin x -1) used the chain rule. f ''(x) = 2cos x * -cos x + -2sin x (-sin x -1) used the product rule. Find the intervals on which f is increasing of decreasing. Increasing if f '(x) > 0. 2cos x < 0 if pi/2
Derivative and the graph of a trig. function?
A couple of ways 1. sin x = cos x => tan x =1 => x = pi/4 and 5pi/4 2. Square both side and get cos ^2 x - sin^2 x = 0 => cos 2x = 0 => 2x = pi/2 , 3pi/2, 5pi/2, 7pi/2 => x = pi/4, 3pi/4, 5pi/4, 7pi/4. Checking the solutions, 3pi/4 and 7pi/4 are not acceptable (They came around when squaring). Also, you should memorize the value of the trig functions at 0, pi/6, pi/3, pi/4, pi/2 etc. You would have just written down the answers pi/4 and 5pi/4. Next, find out where cos x is > sin x. That says that the derivative is positive, hence the function is increasing. Do the same for cos x < sin x: there the function decreases. I'll work the first solution, x = pi/4 For x < pi/4, the cos is greater than the sin => the function increases For x > pi/4, the cos is less than the sin => the function decreases So the function first increases then decreases => it's a maximum
How do I find the range of sin^4x+cos^4x?
[math]\sin^4(x) + \cos^4(x)= (\sin^2(x) + \cos^2(x))^2 - 2 \sin^2(x). \cos^2(x)[/math][math] =1-\frac{1}{2} \sin^2(2x)[/math]So it simplifies to the above expression. Now, range of [math]\sin^2(2x)[/math] is [math][0,1][/math]Therefore, range of [math]1/2(\sin^2(2x)) [/math] is [math][0,1/2][/math].Therefore, range of [math]1-\frac{1}{2} \sin^2(2x)[/math] is [math][1/2,1][/math].As simple as that.... If you didn't get something don't hesitate to leave a comment :D