Find the derivative.

Question

Find the derivative of the function f for n = 1, 2, 3, and 4.

f(x) = cos(x)

x^n

Use the results to write a general rule for f ′(x) in terms of n.

f'(x) =

Answer 1

If f(x)= cosx / x, then f'(x) = (-xsinx - cosx )/ x²

If f(x) = cosx / x², then f'(x) = (-x²sinx - 2xcosx) / x⁴

If f(x) = cosx / x³, then f'(x) = (-x³sinx - 3x²cosx) / x⁶, etc

For f(x) = cosx / xⁿ, where n = 1, 2, 3, 4, ..., f'(x) = (-xⁿsinx - nx^n-1cosx) / x²ⁿ

Answer 2

f(x) = cosx/x^n

f'(x) = cosx)(x^-n)

= (cosx)(-nx^(-n-1)) + (-sinx)(x^-n)

f'(x) = -ncosx/x^(n+1) - sinx/x^n

f'(1) = -cosx/x^2 -sinx/x = -(cosx+xsinx)/x^2

f'(2)= -cosx/x^3 -sinx/x^2 =-(cosx+x^2sinx)/x^3

f'(3) =-cosx/x^4 -sinx/x^3=-(cosx +x^3sinx)/x^4

f'(4)=-cosx/x^5 -sinx/x^4= -(cosx +x^4sinx)/x^5

f'(x) = -(cosx +(x^n)(sinx))/x^(n+1)

UNLESS your problem was really designed to find a formula for fⁿ(x) where f(x) =cos(x) and fⁿ(x) is the nth derivative

then

f'(x) = -sinx where n=1

f"(x) = -cosx where n=2

f"'(x) =sinx where n=3

f^iv(x) =cosx where n=4

general formula is

fⁿ(x) = cosx when n is divisible evenly by 4

but =-sinx when n divided by 4 has a remainder of 1

and = -cosx when n divided by 4 has a remainder of 2

and = sinx when n divided by 4 has a remainder of 3