This function is comprised of two simple functions multiplied together, so we'll use product rule: the derivative of [f(x) * g(x)] = f(x)*g'(x) + f'(x)*g(x).
In this case,
f(x) = sin(x)
which means f'(x) = cos(x)
g(x) = cos(x)
which implies g'(x) = -sin(x)
So, following the formula above, the derivative of sin(x)*cos(x) is
= sin(x)*-sin(x) + cos(x)*cos(x)
= -sin2(x) + cos2(x)
= cos(2x)
This last line is one of the many trig identities that are hard to remember.
(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).
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) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
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.
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is 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).
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
The derivative with respect to 'x' of sin(pi x) ispi cos(pi 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).
sin integral is -cos This is so because the derivative of cos x = -sin x
y = x sin(x) + cos(x)Derivative of the first term = x cos(x) + sin(x)Derivative of the second term = -sin(x)y' = Sum of the derivatives = x cos(x) + sin(x) - sin(x)= [ x cos(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))
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.
0.5*cos(x)/sqrt(sin(x))
It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.