The derivative with respect to 'x' of sin(pi x) ispi cos(pi x)
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
d/dx [sin(x) + 2] = cos(x)
d/dx(-cosx)=--sinx=sinx
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
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))
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = 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))