# Find the derivative. The chain rule.

• Nov 2nd 2013, 03:33 PM
sepoto
Find the derivative. The chain rule.
I am being asked to find the derivative by what I think is the chain rule of this equation:

$(x^2+2)^2$

Let u be:

$u=(x^2+2)^2$

This is the part I am having trouble understanding. The formula for the derivative of $u^2$ is given as:

$\frac{d}{dx}u^2=\frac{du}{dx}\frac{d}{du}u^2=(2x)( 2u)=4x(x^2+2)$

Is that the chain rule? How was $(2x)(2u)$ calculated?

• Nov 2nd 2013, 03:40 PM
Plato
Re: Find the derivative. The chain rule.
Quote:

Originally Posted by sepoto
I am being asked to find the derivative by what I think is the chain rule of this equation:
$(x+2)^2$.

Why do you complicate things?
$\frac{d}{{dx}}{\left( {x + 2} \right)^2} = 2\left( {x + 2} \right)$
• Nov 5th 2013, 01:23 PM
SlipEternal
Re: Find the derivative. The chain rule.
If you want to use substitution, let $u = x^2+2$ and then $(x^2+2)^2 = u^2$ and $\dfrac{du}{dx} = 2x$. So,

$\dfrac{d}{dx}(x^2+2)^2 = \dfrac{d}{dx}(u^2)$.

Then we apply the Chain Rule:

$\dfrac{d}{dx}(u^2) = \dfrac{d}{du}(u^2)\dfrac{du}{dx}$

Notice that on the RHS, you are now differentiating $u^2$ with respect to $u$ (which is what you want).

$\dfrac{d}{du}(u^2)\dfrac{du}{dx} = (2u)(2x)$

Now, we plug back in for $u$ from our substitution:

$(2u)(2x) = 2(x^2+2)(2x) = 4x(x^2+2)$

So, $\dfrac{d}{dx}(x^2+2)^2 = 4x(x^2+2)$
• Nov 5th 2013, 01:44 PM
HallsofIvy
Re: Find the derivative. The chain rule.
Quote:

Originally Posted by sepoto
I am being asked to find the derivative by what I think is the chain rule of this equation:

$(x^2+2)^2$

Let u be:

$u=(x^2+2)^2$

No. $u= x^2+ 2$ but that may be a typo.

Quote:

This is the part I am having trouble understanding. The formula for the derivative of $u^2$ is given as:

$\frac{d}{dx}u^2=\frac{du}{dx}\frac{d}{du}u^2=(2x)( 2u)=4x(x^2+2)$

Is that the chain rule? How was $(2x)(2u)$ calculated?

Yes, that is the chain rule: if u is a function of f, then $\dfrac{df(u(x))}{dx}= \dfrac{df}{du}\dfrac{du}{dx}$.
Here, $f(u)= u^2$ so $\dfrac{df}{du}= 2u$ and $\dfrac{du}{dx}= 2x$.
Then $\dfrac{df}{dx}= (2u)(2x)= (2(x^2+ 2))(2x)= 4x(x^2+ 2)$.