The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals 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))
The derivative of sin(x) is cos(x).
The derivative of y = sin(3x + 5) is 3cos(3x + 5) but only if x is measured in radians.
d/dx (cos x)^2 using the rule of chain, take derivative of the external, times derivative of the internal = 2 (cos x)(-sin x) =-2sinx cos x = - sin(2x)
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is 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).
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
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
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))
Derivative of x = 1, and since sqrt(x) = x^(1/2), derivative of x^(1/2) = (1/2)*(x^(-1/2))Add these two terms together and derivative = 1 + 1/(2*sqrt(x))
For example. d/dx sin^-1 X = 1/sqrt(1 - x^2) Probably derived from the Pythagorean theorem.
sqrt(x) = x^(1/2) The derivative is (1 / 2) * x^(-1 / 2) = 1 / (2 * x^(1 / 2)) = 1 / (2 * sqrt(x))
The formula for the derivative of an inverse (finv)' = 1/(f' o (finv)) allows you get a formula for the derivative of the inverse of any function that you already know the derivative of. For example: What is the derivative of sqrt(x)? You could figure this out using the definition of the derivative, but it is complicated. You already know that the derivative of x2 is 2x. So let f = x2; finv = sqrt(x), f' = 2x. This gives: (sqrt(x))' = 1/(2 sqrt(x)). Now you have derived a "square root rule" with almost no work.