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f(x) = 4.1cos9t
the derivitative of just cos(9t) is -9sin9t
then just multiply -9 by 4.1 (-36.9)
the derivative is -36.9sin(9t)
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What is the derivative of cos x sin x divided by cos x sin x?
(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
What is the second derivitive of sec x?
Write sec x as a function of sines and cosines (in this case, sec x = 1 / cos x). Then use the division formula to take the first derivative. Take the derivative of the first derivative to get the second derivative. Reminder: the derivative of sin x is cos x; the derivative of cos x is - sin x.
How do you differentiate cosine square root of x?
The derivative of cos x is -sin x, the derivative of square root of x is 1/(2 root(x)). Applying the chain rule, the derivative of cos root(x) is -sin x times 1/(2 root(x)), or - sin x / (2 root x).
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.
Proof of the derivative of the cosecant function?
Express the cosecant in terms of sines and cosines; in this case, csc x = 1 / sin x. This can also be written as (sin x)-1. Remember that the derivative of sin x is cos x, and use either the formula for the derivative of a quotient (using the first expression), or the formula for the derivative of a power (using the second expression).
What is the derivative of ln cos x?
The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).
What is the derivative of cos x sin x divided by cos x sin x?
(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
What do derivative classifiers do?
Cos-1 2x2-1
What must Derivative classifiers do?
Cos-1 2x2-1
Derivative classifiers must what?
Cos-1 2x2-1
What is the second derivitive of sec x?
Write sec x as a function of sines and cosines (in this case, sec x = 1 / cos x). Then use the division formula to take the first derivative. Take the derivative of the first derivative to get the second derivative. Reminder: the derivative of sin x is cos x; the derivative of cos x is - sin x.
What is the derivative of the square root of cos x?
To find the derivatve of the square root of cos x: Use the chain rule; this means multiply the inner derivative by the outer derivative. You can write the question f(x) = (cos x)1/2 This general break-down explains how to find d/dx f(x) note: d/dx basically symbolizes "the derivative of" In general terms: f(x) = x1/2 g(x) = cos x f(g(x)) = (cos x)1/2 outer derivative: d/dx f(z) = (1/2)*x-1/2 = 1/(4cos x)1/2 inner derivative: d/dx g(x) = -sin(x) final answer: d/dx f(g(x)) = -sin(x)/(4*cos x)1/2 note: d/dx means "the derivative of"; so d/dx x = 1 Further explained: Set up the equation to a more general form: (cos x)1/2 To make the inner derivative, look at cos(x) To make the outer derivative, look at x1/2 note: x ~ cos x; so we treat (cos x) simply as x, to create the outer derivative You probably know the necessary derivates: 1. derivative of cos x = -sin x 2. derivative of a1/2 = (1/2)*a-1/2 = 1/(4a)1/2 Multiplying the two we get the answer: -sin(x)/(4cos x)1/2
What is the derivative of cos-13 5cosx5 3cosx?
d(cos−13 + 5cosx5 + 3cosx)/dx = −1/2(1−32) − 25x4sinx5 − 3sinx
How do you differentiate cosine square root of x?
The derivative of cos x is -sin x, the derivative of square root of x is 1/(2 root(x)). Applying the chain rule, the derivative of cos root(x) is -sin x times 1/(2 root(x)), or - sin x / (2 root x).
How do you factor 9t plus 9?
9t + 9 = 9 (t + 1)
What is the antiderivative of sqrtx plus 4ex minus sinx?
The anti-derivative of sqrt(x) : sqrt(x)=x^(1/2) The anti-derivative is x^(1/2+1) /(1/2+1) = (2/3) x^(3/2) The anti-derivative is 4e^x is 4 e^x ( I hope you meant e to the power x) The anti-derivative of -sin(x) is cos(x) Adding, the anti-derivative is (2/3) x^(3/2) + 4 e^x + cos(x) + C
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.
