How do you use the product Rule to find the derivative of (10-2x)(6-2x)(x) + x^3?

Aug 10, 2015

${y}^{'} = 15 {x}^{2} - 64 x + 60$

Explanation:

Your function can be written as the sum of two functions, let's say $f \left(x\right)$ and $g \left(x\right)$

$y = f \left(x\right) + g \left(x\right)$

where $f \left(x\right) = \left(10 - 2 x\right) \left(6 - 2 x\right) \cdot x$ and $g \left(x\right) = {x}^{3}$.

This means that the derivative of $y$ will take the form

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$

Now, you can differentiate $f \left(x\right)$, which can be written as the product of three other functions, by using the product rule.

For a function $f \left(x\right)$ that can be written as

$f \left(x\right) = h \left(x\right) \cdot i \left(x\right) \cdot k \left(x\right)$

you can find its derivative by using the formula

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(h \left(x\right)\right)\right] \cdot i \left(x\right) \cdot k \left(x\right) + h \left(x\right) \cdot \left[\frac{d}{\mathrm{dx}} \left(i \left(x\right)\right)\right] \cdot k \left(x\right) + h \left(x\right) \cdot i \left(x\right) \cdot \left[\frac{d}{\mathrm{dx}} \left(k \left(x\right)\right)\right]}$

In your case, you have

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(10 - 2 x\right)\right] \cdot \left(6 - 2 x\right) \cdot x + \left(10 - 2 x\right) \cdot \left[\frac{d}{\mathrm{dx}} \left(6 - 2 x\right)\right] \cdot x + \left(10 - 2 x\right) \cdot \left(6 - 2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(x\right)$

${f}^{'} = \left(- 2\right) \cdot \left(6 - 2 x\right) \cdot x + \left(10 - 2 x\right) \cdot \left(- 2\right) \cdot x + \left(10 - 2 x\right) \cdot \left(6 - 2 x\right) \cdot 1$

${f}^{'} = - 12 x + 4 {x}^{2} - 20 x + 4 {x}^{2} + 60 - 32 x + 4 {x}^{2}$

${f}^{'} = 12 {x}^{2} - 64 x + 60$

Your target derivative will thus be

${y}^{'} = 12 {x}^{2} - 64 x + 60 + \frac{d}{\mathrm{dx}} \left(3 {x}^{2}\right)$

${y}^{'} = 12 {x}^{2} - 64 x + 60 + 3 {x}^{2}$

${y}^{'} = \textcolor{g r e e n}{15 {x}^{2} - 64 x + 60}$