# Thread: product rule, taking derivatives

1. ## product rule, taking derivatives

2. Find the derivative of $y = 3 sin^{3} (2x^{4} + 1)$

I used the product rule and this is my working, however I don't end up with the correct answer, can someone please tell me where I am going wrong, thanks.

$u = 3sin^{3}$

$v = 2x^{4} +1$

$t = sin , y= 3t^{3}$

$\frac{dt}{dx} = cos$

$\frac{dy}{dt} = 9t^{2}$

so now I multiply both together to get the derivative of $3sin^{3}$

$9sin^{2} \times cos$

$\frac{dy}{dx} = 8x^{3}$

Now using the product rule and putting this together.

$9sin^{2} \times cos \times (2x^{4} +1) + 8x^{3} \times 3sin^{3}$

$72x^{3} sin^{2}(2x^{4} +1)cos(2x^{4} +1)$

2. not the product rule ... the chain rule

$y = 3[\sin(2x^4+1)]^3$

$
y' = 9[\sin(2x^4+1)]^2 \cdot \cos(2x^4+1) \cdot 8x^3
$

now clean up the algebra

3. Originally Posted by skeeter
not the product rule ... the chain rule

$y = 3[\sin(2x^4+1)]^3$

$
y' = 9[\sin(2x^4+1)]^2 \cdot \cos(2x^4+1) \cdot 8x^3
$

now clean up the algebra
Oh I see.

But I am just wondering how did you know to use the chain rule instead of the product rule? As this expression is a product of two functions, so that's why I assumed I should use the product rule.

and I thought the chain rules was supposed to be used for a function of a function. Like $\sqrt{3x-1}$

So for an expression like this $x^{2} \sqrt{3x-1}$

would you use the chain rule or product rule?

thank you for clarifying, greatly appreciated.

4. Originally Posted by Tweety
Oh I see.

But I am just wondering how did you know to use the chain rule instead of the product rule? As this expression is a product of two functions, <<<< No. See below
so that's why I assumed I should use the product rule.

and I thought the chain rules was supposed to be used for a function of a function. <<<<< Correct

...
Your function is indeed the function of a function of a function:

$f_1(x)=2x^4+1$

$f_2(x)=\sin(x)$

$f_3(x)=3x^3$

$f(x)=f_3(f_2(f_1(x)))$
$\sin^2(x)=(\sin(x))^2$
$\sin^2(x) \neq \sin((x)^2)$