Differentiation and Chain Rule

**madmax29** · November 7th 2010, 03:31 AM

Question: y=3Sin^4(2x)

...how do you solve this using the chain rule;

dy/dx= dy/du X du/dx,

y =3Sin^4 where u = 2x <<<< is this correct?

what do i do after this, can someone explain

i think im using differentiation rules;
sin x = cos x
x^n = nx ^n-1

**e^(i*pi)** · November 7th 2010, 03:42 AM

Yes, although you need to use the chain rule three times because of the sin function

Set $u = 2x$ and $v = sin(u)$ so that $y=3v^4$

By the chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{dv} \cdot \dfrac{dv}{du} \cdot \dfrac{du}{dx}$

**madmax29** · November 7th 2010, 03:58 AM

therefore

dy/du= nx^n-1 >>> = 3v^4

du/dv= 2x >>> =2

dv/dx= sin (u) >>> cos(u)

therefore

$\dfrac{dy}{dx} = \dfrac{dy}{dv} \cdot \dfrac{dv}{du} \cdot \dfrac{du}{dx}$

= 3v^3 X Cos (u) X 2

= 2 Cos (2x) 3 Sin^3 (2x)
is this correct????

**e^(i*pi)** · November 7th 2010, 04:05 AM

No, I don't know where your 3 has gone?

Your individual differentation is fine, apart from dy/dv which is 12v^3 (you didn't multiply by that 4)

$12v^3 \cdot \cos(u) \cdot 2 = 24v^3\, \cos(u)$

When we put in our original substitutions back we get:

$24\sin^3(u) \cdot \cos(u) = 24\sin^3(2x)\cos(2x)$

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