what is the derivative of PI?

Question

The derivative of PI is ZERO, right?

so the derivative of PI(cos(x)) would equal -sin(x)?

Answer 1

The derivative of a constant is 0 and pi is a constant.



The derivative of a constant times a function is the constant multiplied by the derivative of the function.



derivative of pi*cos(x) = pi * derivative of cos(x) = -pi*sin(x)

Answer 2

Yes, pi is a constant, so its derivative is zero.



However, the derivative of n * f(x) with respect to x, where n is a constant, is n * derivative of f(x) with respect to x. The constant remains in the derivative. REMEMBER THIS!



So, the derivative of pi (cos (x)) is pi * derivative of cos x = pi * -sin x = -pi sin x.



I hope that helps. :)



ILoveMaths07.

Answer 3

f(x) = (pi)/x^2 Isolate the coeficient pi So you have (pi) x derivative of 1/x 1/x >> x^-1 >> derivative of x^-1 is -x^-2 >> 1/x^2 So it is (pi)(-1/x^2) For the integral is: You have to cancel the x^2. For this you can visualize that if you take the U of cos, you can cancel the x^2 of the bottom. You take as U pi/x. You already took the derivative of pi/x... so du = (pi)(-1/x^2) dx dx = 1/(pi)(-1/x^2) dx = (-x^2)/pi So it is [(-x^2)/pi] cosU / (x^2) x^2 get cancelled And you have (-1/pi) cosU which is: (-1/pi) cos(pi/x) Remember integral of cos is (sin): So the final answer is: - (sin(pi/x)) / pi

Answer 4

No, the derivative of pi*cos(x) is -pi*sin(x). Use the product rule.

Answer 5

Pi is a constant so you would treat it as if it were just a number out front

it would be

-pi*sin(x)

Answer 6

No, When getting the derivative of Trigonometric functions such as sin and cos the scalar factor remains such as



The derivative of 3cosx is -3sinx

or derivative of 5sinx is 5cosx



Thus since π is just a number it remains:



The derivative of π(cos(x)) is -π(sin(x))

Answer 7

not really - PI is a constant, so derivative of PI alone is zero, but how would you treat the derivative of 2X? so the aswer should be -PIsin(x)

Answer 8

We know derivatives of π and cos(x):

d/dx (π) = 0

d/dx (cos(x)) = -sin(x)



So what is derivative of π*cos(x)?

You can't just add the derivatives or multiply them.

Use product rule:

derivative of f(x)g(x) = f'(x)g(x) + f(x)g'(x)



d/dx (πcos(x))

= d/dx (π) * cos(x) + π * d/dx (cos(x))

= 0 cos(x) + π (-sin(x))

= - π sin(x)



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Note: I just used this method as a demonstration.

When f(x) is a constant, it is not necessary to find f'(x)g(x), since f'(x) = 0, so this term is also 0.

We are just left with f(x)*g'(x), i.e. the constant (π) * derivative of g(x) which in this case is cos(x)