Derivatives...?

Question

If I'm not mistaken, the derivative of f(x)=sin(x)cos(x) is -sin^2(x)+cos^2(x) using the formula f'(x)g(x)+f(x)g'(x). In my book the answer to the second and third derivatives are -4sin(x)cos(x), 4sin^2(x)-4cos^2(x) respectably. How do they get that from the first derivative?

Answer 1

f(x)=sinxcosx=1/2(2sinxcosx)
=(1/2)sin2x
f'(x)=1/2*2*cos2x
=cos2x
f"(x)=-2sin2x=-4sinxcosx
f'''(x)=4sin^2(x)-4cos^2(x)

Answer 2

because the derivative of sin^2(x) by that product rule formula equals 2*sin(x)*[-cos(x)]. So does the derivative of cos^2(x) when you do it out.

Add them up to get -4sin(x)cos(x) for the second derivative.

to find the third derivative we can employ a little shortcut if we want. It's this: the second derivative equals -4 times the original function, so the derivative of the second derivative (ie the third derivative) will equal -4 times the first derivative, ie 4*(sin^2(x)-cos^2(x)). hopefully that's clear.

Answer 3

The first derivative of f(x) = sinx*cosx is, like you said, f'(x) = -sin²x + cos²x.

To get the second derivative, you take the derivative of the first derivative. You can use the power rule like you do with the regular functions.

f'(x) = -sin²x + cos²x
f"(x) = -2sinx*cosx + 2cosx*(-sinx)
f"(x) = -4sinx*cosx

And then to get the third derivative you take the derivative of the second derivative. This time you need to use the product rule.

Answer 4

your first derivative is correct.

f'(x) = -sin^2(x)+cos^2(x)
use the chain rule and you end up with

f''(x) = -2sin(x)cos(x)+2cos(x)(-sin(x)
f''(x)= -4sin(x)cos(x)

the third derivative is the same as the first derivative but with a coeffectient of -4.

Answer 5

You could notice that sinx*cosx = 1/2*sin(2x).

Your derivative, which is correct, is the formula for cos(2x). When you take the derivative of cos(2x) you get - tada! -2sin2x = -4sinxcosx

Answer 6

as you say f'(x) =cos^2(x) -sin^2(x)

using function of function and product rule

f''(x) = 2cos(x) [- sin(x) ] - 2sin(x) [cos(x) ] = -2cos(x) sin(x)

Using product rule

f'''(x) = -2 [cos(x) *cos(x) + sin(x) *{-sin(x) } ]

= 4sin^2(x)-4cos^2(x)


but it is easier to use trigonometric formulae as show in some of the other answers

Answer 7

easiest way is to notice that, cos^2(x)-sin^2(x) = cos(2x)

Answer 8

By using the power formula
y = f(x)^n => dy/dx = nf(x)^(n-1)*f'(x) and a bit of algebra.


Doug