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The derivative of cot(x) is -csc2(x).
(Which is the same as -1/sin2(x).)
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What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
What is derivative of cot t dt?
d/dt cot (t) dt = - cosec2(t)
How do you draw the graph of modulus of y equals cot x?
First note that this not the graph of y = |cot(x)|.The equivalent equations for |y| = cot(x) or cot(x) = |y| arecot(x) = -y or cot(x) = +ySo plot y = cot x and then reflect all the points in the x-axis.
What is the derivative of 2lnx?
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
How do you simplify sec x cot x?
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)
What is the derivative of cotx?
The derivative of cot(x) is -csc2(x).
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
What is the derivative of cot?
d/dx (cot x) = -csc2x
Second derivative of cosecx?
d/dx cosec(x) = - cosec(x) * cot(x) so the second derivative or d(d/dx)/dx cosec(x) = [- cosec(x) * d/dx cot(x)] + [ - d/dx cosec(x) * cot(x)] = [- cosec(x) * -cosec^2(x)] + [ - (- cosec(x) * cot(x)) * cot(x)] = cosec(x) * cosec^2(x) + cosec(x)*cot^2(x) = cosec(x) * [cosec^2(x) + cot^2(x)].
What is the anti-derivative of co secant x?
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
What is the derivative of 3tanx-4cscx?
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What is the second derivative of ln(tan(x))?
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2
What is derivative of cot t dt?
d/dt cot (t) dt = - cosec2(t)
What is the derivative of x equals cot2 divided by t?
Assuming you want dx/dt and that the equation is x = cot(2) / t (i.e. cot(2) is a constant) we can use the power rule. First, we rewrite it: cot(2)/t = cot(2) * t-1 thus, by the power rule: dx/dt = (-1) cot(2) * t-1 -1 = - cot(2) * t-2= = -cot(2)/t2
What is the derivative of sinx pwr cosx?
For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.
How do you draw the graph of modulus of y equals cot x?
First note that this not the graph of y = |cot(x)|.The equivalent equations for |y| = cot(x) or cot(x) = |y| arecot(x) = -y or cot(x) = +ySo plot y = cot x and then reflect all the points in the x-axis.
What is cot x?
Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1
