How do you differentiate f(x)= (5x+4x^4)(5+3x^2)  using the product rule?

Mar 23, 2018

The derivative of $f \left(x\right)$ is $72 {x}^{5} + 80 {x}^{3} + 45 {x}^{2} + 25$.

Explanation:

Here's the product rule:

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

An easy way to remember this is to say "Left d-Right, Right d-Left", meaning the product is the left function times the derivative of the right, plus the right function times the derivative of the left.

Here's the product rule applied to our problem:

$\textcolor{w h i t e}{=} \frac{d}{\mathrm{dx}} \left(5 x + 4 {x}^{4}\right) \left(5 + 3 {x}^{2}\right)$

$= \left(5 x + 4 {x}^{4}\right) \cdot \frac{d}{\mathrm{dx}} \left(5 + 3 {x}^{2}\right) + \left(5 + 3 {x}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left(5 x + 4 {x}^{4}\right)$

$= \left(5 x + 4 {x}^{4}\right) \setminus \cdot \left(0 + 6 x\right) + \left(5 + 3 {x}^{2}\right) \cdot \left(5 + 16 {x}^{3}\right)$

$= \left(5 x + 4 {x}^{4}\right) \setminus \cdot 6 x + \left(5 + 3 {x}^{2}\right) \cdot \left(5 + 16 {x}^{3}\right)$

$= 30 {x}^{2} + 24 {x}^{5} + \left(5 + 3 {x}^{2}\right) \cdot \left(5 + 16 {x}^{3}\right)$

$= 30 {x}^{2} + 24 {x}^{5} + 25 + 80 {x}^{3} + 15 {x}^{2} + 48 {x}^{5}$

$= 24 {x}^{5} + 25 + 80 {x}^{3} + 45 {x}^{2} + 48 {x}^{5}$

$= 72 {x}^{5} + 25 + 80 {x}^{3} + 45 {x}^{2}$

$= 72 {x}^{5} + 80 {x}^{3} + 45 {x}^{2} + 25$

That's the derivative. Hope this helped!