Differentiate y=x sinx cosx

[zonniboy]

I'm not sure how to do this one. Only way I can think of is using the Product Rule but I don't know how to apply it when there are more than two functions.

Something like:

y=(x)(sinx)(cosx)

--separate it into three different functions--

f(x)=x, g(x)=sinx, z(x)=cosx

--use Product Rule--

???

Are those the right steps to differentiate the function and if they are how do I apply the Product Rule to three functions instead of just two?

Any help would be appreciated.

[Hurkyl]

whats wrong with

f(x) = x
g(x) = (sin x)(cos x)

?

[zonniboy]

Ah. Didn't think of it that way. Sorry, brand new to Trig and Calc.

[Hurkyl]

There is a generalization of the product rule; I just wanted to realize this other approach to the problem because this idea it may be useful in the future. (and, IMHO, is easier to remember!)

Anyways, the generalization is:


\begin{array}{l}
(f_1 f_2 f_3 \ldots f_n)' = \\
f_1' f_2 f_3 \ldots f_n +
f_1 f_2' f_3 \ldots f_n +
f_1 f_2 f_3' \ldots f_n +
\ldots +
f_1 f_2 f_3 \ldots f_n'
\end{array}

[Orion1]

Differentiate:
y = x(sin(x))(cos(x))

f(x) = x
g(x) = sin(x)
h(x) = cos(x)

Product Rule:
(d/dx)[f(x)g(x)h(x)] = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)

Factor g(x):
(d/dx)[f(x)g(x)h(x)] = f(x)h(x)g'(x) + g(x)(h(x)f'(x) + f(x)h'(x))
(d/dx)[f(x)g(x)h(x)] = x*cos(x)*(d/dx)[sin(x)] + sin(x)(cos(x)(d/dx)[x] + x(d/dx)[cos(x)])

Derivative:
(d/dx)[f(x)] = (d/dx)[x] = 1
(d/dx)[g(x)] = (d/dx)[sin(x)] = cos(x)
(d/dx)[h(x)] = (d/dx)[cos(x)] = -sin(x)

= x*cos(x)*cos(x) + sin(x)(cos(x)(1) + x*-sin(x))
= x*cos^2(x) + sin(x)(cos(x) - x*sin(x))
= x*cos^2(x) + sin(x)*cos(x) - x*sin^2(x)
= sin(x)*cos(x) + x*cos^2(x) - x*sin^2(x)

Factor x:
= cos(x)sin(x) + x(cos^2(x) - sin^2(x))
x(cos^2(x) - sin^2(x)) = x*cos(2x)

Solution:
= x*cos(2x) + cos(x)*sin(x)

---

(d/dx)[f(x)g(x)h(x)] = (d/dx)[x(sin(x))(cos(x))]

Use Product Rule:
\frac{d(u,v)}{dx} = \frac{du}{dx}v + u \frac{dv}{dx}
u = x, v = cos(x)sin(x)
(d/dx)[x] = 1
= x(d/dx)[cos(x)sin(x)] + cos(x)sin(x)(d/dx)[x]
= x(d/dx)[cos(x)sin(x)] + cos(x)sin(x)

Use Product Rule:
\frac{d(u,v)}{dx} = \frac{du}{dx}v + u \frac{dv}{dx}
u = cos(x), v = sin(x)
= cos(x)sin(x) + x(cos(x)(d/dx)[sin(x)] + sin(x)(d/dx)[cos(x)])
(d/dx)[sin(x)] = cos(x)
(d/dx)[cos(x)] = -sin(x)
= cos(x)sin(x) + x(cos^2(x) - sin^2(x))
x(cos^2(x) - sin^2(x)) = x*cos(2x)

Solution:
= x*cos(2x) + cos(x)sin(x)

[h2]

xcosxsinx = xsin(2x)/2 so
(xcosxsinx)' = sin(2x)/2 + xcos(2x)

[Chen]

As h2 said, use the identity.

f(x) = x\cos x\sin x = \frac{x\sin 2x}{2}
f'(x) = \frac{1}{2}(\sin 2x + 2x\cos 2x) = \frac{\sin 2x}{2} + x\cos 2x = \cos x\sin x + x\cos 2x