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y = 2sin(x)cos(x)
Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:
y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)
Simplify:
y' = 2cos2(x)-2sin2(x)
y' = 2(cos2(x)-sin2(x))
Use trig identity cos(2x) = cos2(x)-sin2(x):
y' = 2cos(2x)
Take second derivative using chain rule:
y'' = 2(-sin(2x)cos(2x))
Simplify:
y'' = -2sin(2x)(2)
Simplify:
y'' = -4sin(2x)y'' = -4sin(2x)
Add your answer:
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What is the differentiation of sin x over 1 plus cosx?
The derivative is 1/(1 + cosx)
What is the derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
How do you find the solutions of tanx equals 2cscx?
tanx=2cscx sinx/cosx=2/sinx sin2x/cosx=2 sin2x=2cosx 1-cos2x=2cosx 0=cos2x+2cosx-1 Quadratic formula: cosx=(-2±√(2^2+4))/2 cosx=(-2±√8)/2 cosx=(-2±2√2)/2 cosx=-1±√2 cosx=approximately -2.41 or approximately 0.41. Since the range of the cosine function is [-1,1], only approx. 0.41 works. So: cosx= approx. 0.41 Need calculator now (I went as far as I could without one!) x=approx 1.148
How do you differentiate sin sin x?
To differentiate y=sin(sin(x)) you need to use the chain rule. A common way to remember the chain rule is "derivative of the outside, keep the inside, derivative of the inside". First, you take the derivative of the outside. The derivative of sin is cos. Then, you keep the inside, so you keep sin(x). Then, you multiple by the derivative of the inside. Again, the derivative of sinx is cosx. In the end, you get y'=cos(sin(x))cos(x))
What is sin 3x in terms of sin x?
given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cos²x.sinx - sin³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x
What is the anti-derivative of -2cosx?
take out the constant -2 then take the intergral of cosx this will give you sinx your answer is -2sinx
How would you find x when 0 equals 2sinxcosx-cosx?
2sinxcosx-cosx=0 Factored : cosx(2sinx-1)=0 2 solutions: cosx=0 or sinx=.5 For cosx=0, x=90 or 270 degrees For sinx=.5, x=30 degrees x = {30, 90, 270}
What is the derivative of -cosx?
-sinx
What is the derivative of x-cosx?
Take the derivative term by term. d/dx(X - cosX) = sin(X) ======
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
What is the derivative of sin x minus cos x?
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
What is the differentiation of sin x over 1 plus cosx?
The derivative is 1/(1 + cosx)
What is the derivative of the square root of 2 raised to the power of cos x?
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
What is the derivative of 7 to the power of cosx?
derivative (7cosx) = -ln(7) 7cosx sinx dx
Derivative of cosx?
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(x).
Find the derivative of sinxcosx?
(cosx)^2-(sinx)^2
What is the derivative of 1 plus cosx?
negative sin(x)
