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If fx equals cossinx2 then f prime equals?
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
If y equals -2 when x equals 4 find x when y equals 5?
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
What number is needed to complete the square x2 23x equals 0?
x2 23x equals 0
How do you solve x2 equals cos x?
You can solve this to the accuracy of your liking by using Newton's method: xn+1 = xn - f(xn) / f'(xn) In this case, we'll say f(x) = x2 - cos(x) f'(x) would then be 2x + sin(x) Let's take a rough guess, and start with x0 = 0.5 x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536 x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916 x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712 x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229 x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297 x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296 Now we can test our answer: 0.824132312302522422962 = 0.67919406818110235182 cos(0.82413231230252242296) = 0.67919406818110235183 So we're accurate to the nearest ten quintillionth.
If c equals x2 then what is Cx2?
if c = x2 then cx2 = x4
If fx equals cossinx2 then f prime equals?
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
What is the derivative of 3cosx2?
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
What does x2 plus x2 equals?
x2 + x2 = 2x2
If y equals -2 when x equals 4 find x when y equals 5?
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
What number is needed to complete the square x2 23x equals 0?
x2 23x equals 0
Finf f prime x and g prime x if f x equals sin of 1 over x and g x equals 1 over sin x?
f(x)= sin(1/x) and g(x)=1/sin(x) [u(v)]' = u'(v) * v', where u and v are functions So f'(x) = sin'(1/x) * (1/x)' = cos(x) * (-1/x2) = -cos(x)/x2 g'(x) = (1/x)' applied to sin(x) * (sin(x))' = -1/(sin2(x)) * cos(x) = -cos(x)/(sin2(x))
What does x2 plus 6-40 equal?
It equals x2 - 34
Solve x2 equals 64?
x2 = 6482 = 64x = 8
How do you isolate x2 10x2 - 36y2 equals 100?
x2 − 36y2
If c equals x2 then what is Cx2?
if c = x2 then cx2 = x4
Which values are solution to the inequality of x2 equals 64?
x2≤64
How do you solve x2 equals cos x?
You can solve this to the accuracy of your liking by using Newton's method: xn+1 = xn - f(xn) / f'(xn) In this case, we'll say f(x) = x2 - cos(x) f'(x) would then be 2x + sin(x) Let's take a rough guess, and start with x0 = 0.5 x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536 x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916 x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712 x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229 x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297 x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296 Now we can test our answer: 0.824132312302522422962 = 0.67919406818110235182 cos(0.82413231230252242296) = 0.67919406818110235183 So we're accurate to the nearest ten quintillionth.
