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:

$\displaystyle (x^2+2)^2$

Let u be:

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

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

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

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

Thanks in advance...

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:

$\displaystyle (x+2)^2$.

Why do you complicate things?

$\displaystyle \frac{d}{{dx}}{\left( {x + 2} \right)^2} = 2\left( {x + 2} \right)$

Re: Find the derivative. The chain rule.

If you want to use substitution, let $\displaystyle u = x^2+2$ and then $\displaystyle (x^2+2)^2 = u^2$ and $\displaystyle \dfrac{du}{dx} = 2x$. So,

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

Then we apply the Chain Rule:

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

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

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

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

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

So, $\displaystyle \dfrac{d}{dx}(x^2+2)^2 = 4x(x^2+2)$

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:

$\displaystyle (x^2+2)^2$

Let u be:

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

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

Quote:

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

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

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

Thanks in advance...

Yes, that is the chain rule: if u is a function of f, then $\displaystyle \dfrac{df(u(x))}{dx}= \dfrac{df}{du}\dfrac{du}{dx}$.

Here, $\displaystyle f(u)= u^2$ so $\displaystyle \dfrac{df}{du}= 2u$ and $\displaystyle \dfrac{du}{dx}= 2x$.

Then $\displaystyle \dfrac{df}{dx}= (2u)(2x)= (2(x^2+ 2))(2x)= 4x(x^2+ 2)$.