d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
Chat with our AI personalities
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
cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x