How do I use the chain rule to differentiate this function?

hadizainud
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Homework Statement



Differentiate the functions using chain rule. 2(x3 −1)(3x2 +1)4

Homework Equations



Chain Rule = f ' (g(x))g' (x)

The Attempt at a Solution



I don't know how to do using chain rule, but product rule is easier

So using product rule,

= f ' (x) g(x) + f (x)g' (x)
= (2)(x3-1)0(3x2)(3x2+1)4 + 2(x3-1)(4)(3x2+1)3(6x)
= (2)(3x2)(3x2+1)4 + 8(x3-1)(3x2+1)3(6x)

Can anyone show me how to do it, using chain rule?
 
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hadizainud said:

Homework Statement



Differentiate the functions using chain rule. 2(x3 −1)(3x2 +1)4

Homework Equations



Chain Rule = f ' (g(x))g' (x)

The Attempt at a Solution



I don't know how to do using chain rule, but product rule is easier

So using product rule,

= f ' (x) g(x) + f (x)g' (x)
= (2)(x3-1)0(3x[SUPdid]2[/SUP])(3x2+1)4 + 2(x3-1)(4)(3x2+1)3(6x)
= (2)(3x2)(3x2+1)4 + 8(x3-1)(3x2+1)3(6x)

Can anyone show me how to do it, using chain rule?
You did use the chain rule , along with the product rule.

You used the chain rule to differentiate (x3 −1) and (3x2 +1)4 !
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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