# How do you find the derivative of u=(x^2+3x+1)^4?

Mar 14, 2018

You have two options: The first is to expand that polynomial and take the derivative. The other is to do a substitution to get the solution: $\frac{\mathrm{du}}{\mathrm{dx}} = 4 {\left({x}^{2} + 3 x + 1\right)}^{3} \cdot \left(2 x + 3\right)$

#### Explanation:

I'm going to go with the second option, because I'm not a masochist (jokes!)

When you take a derivative of a set of terms in parentheses, you can do a substitution, treating the parentheses as a separate single function. We'll treat the in-parentheses terms like this:

$\left({x}^{2} + 3 x + 1\right) = g \left(x\right)$

so our function now looks like this:

$u = g {\left(x\right)}^{4}$

Now, the derivative is not simply $\frac{\mathrm{du}}{\mathrm{dx}} = 4 g {\left(x\right)}^{3}$, but it is instead $\frac{\mathrm{du}}{\mathrm{dx}} = 4 g {\left(x\right)}^{3} \cdot \frac{\mathrm{dg}}{\mathrm{dx}}$. We must calculate $\frac{\mathrm{dg}}{\mathrm{dx}}$ before arriving at our solution.

if $g \left(x\right) = {x}^{2} + 3 x + 1$ then:

$\frac{\mathrm{dg}}{\mathrm{dx}} = 2 x + 3$

Re-inserting our substitutions, we now arrive at our answer:

$\frac{\mathrm{du}}{\mathrm{dx}} = 4 {\left({x}^{2} + 3 x + 1\right)}^{3} \cdot \left(2 x + 3\right)$

Mar 14, 2018

$4 {\left({x}^{2} + 3 x + 1\right)}^{3} \left(2 x + 3\right)$

#### Explanation:

Use the chain and power rule. Find the derivative of $1 {\left(\ldots .\right)}^{4}$(use the power rule. Then find the derivative of $\left({x}^{2} + 3 x + 1\right)$(use the power rule). then combine the two. $4 {\left({x}^{2} + 3 x + 1\right)}^{3} \left(2 x + 3\right)$

Mar 14, 2018

$\frac{\mathrm{du}}{\mathrm{dx}} = 4 \left(2 x + 3\right) {\left({x}^{2} + 3 x + 1\right)}^{3}$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{Given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$u = {\left({x}^{2} + 3 x + 1\right)}^{4}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 4 {\left({x}^{2} + 3 x + 1\right)}^{3} \times \frac{d}{\mathrm{dx}} \left({x}^{2} + 3 x + 1\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}} = 4 \left(2 x + 3\right) {\left({x}^{2} + 3 x + 1\right)}^{3}$