How do you find the derivative of (1+x+[3x^2])^4?

Jun 27, 2018

$4 \cdot \left(1 + 6 x\right) {\left(1 + x + 3 {x}^{2}\right)}^{3}$

Explanation:

$\left({\left(f \left(x\right)\right)}^{n}\right) \frac{\mathrm{dx}}{\mathrm{dy}} = n \cdot f ' \left(x\right) \cdot f {\left(x\right)}^{n - 1}$

$\left({\left(1 + x + \left[3 {x}^{2}\right]\right)}^{4}\right) \frac{\mathrm{dx}}{\mathrm{dy}} = 4 \cdot \left(1 + 6 x\right) {\left(1 + x + 3 {x}^{2}\right)}^{3}$

Jun 27, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} \to f ' \left(x\right) = 4 \left(1 + 6 x\right) {\left(1 + x + 3 {x}^{2}\right)}^{3}$

Explanation:

Using the 'old fashioned' notation

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

Set $u = 1 + x + 3 {x}^{2} \textcolor{w h i t e}{\text{d}}$ then $\textcolor{w h i t e}{\text{d}} \frac{\mathrm{du}}{\mathrm{dx}} = 1 + 6 x$

Also we have $y = {u}^{4} \textcolor{w h i t e}{\text{d")->color(white)("d}} \frac{\mathrm{dy}}{\mathrm{du}} = 4 {u}^{3}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Really the format $\frac{\mathrm{dy}}{\mathrm{dx}}$ is a fraction and it behaves the same way fractions do. Consequently we can multiply them.

Observe that: $\frac{\mathrm{du}}{\mathrm{dx}} \times \frac{\mathrm{dy}}{\mathrm{du}} \textcolor{w h i t e}{\text{d") =color(white)("d") (du)/(du)xxdy/dx color(white)("d")=color(white)("d}} \frac{\mathrm{dy}}{\mathrm{dx}}$

Using the above:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + 6 x\right) \times 4 {u}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + 6 x\right) \times 4 {\left(1 + x + 3 {x}^{2}\right)}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \to f ' \left(x\right) = 4 \left(1 + 6 x\right) {\left(1 + x + 3 {x}^{2}\right)}^{3}$