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(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
The derivative of sin (x) is cos (x). It does not work the other way around, though. The derivative of cos (x) is -sin (x).
cos(x^2)=cos(x times 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.
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)
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(x).
The derivative with respect to 'x' of sin(pi x) ispi cos(pi 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).
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)
Use the rule for multiplication with a constant - and look up the derivative of "cos x" in a basic table of derivatives. The answer is 3 times the derivative of cos x.
(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
The derivative of sin (x) is cos (x). It does not work the other way around, though. The derivative of cos (x) is -sin (x).
cos(x^2)=cos(x times 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.
Every fourth derivative, you get back to "sin x" - in other words, the 84th derivative of "sin x" is also "sin x". From there, you need to take the derivative 3 more times, getting:85th derivative: cos x86th derivative: -sin x87th derivative: -cos x
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)