How do you use the Product Rule to find the derivative of 2x^4 * 6^(3x)?

Oct 24, 2015

$\frac{d}{d} \left(2 {x}^{4} \cdot {6}^{3 x}\right) = \left(6 {x}^{4}\right) \left({6}^{3 x} \ln 6\right) + {6}^{3 x} \left(8 {x}^{3}\right)$

Explanation:

The product rule states that

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f \left(x\right) \cdot g ' \left(x\right) + g \left(x\right) \cdot f ' \left(x\right)$

In this case we may take our $f \left(x\right) = 2 {x}^{4} \mathmr{and} g \left(x\right) = {6}^{3 x}$ and apply the product rule together with other standard rules of differentiation to get

$\frac{d}{d} \left(2 {x}^{4} \cdot {6}^{3 x}\right) = \left(2 {x}^{4}\right) \left({6}^{3 x} \ln 6 \cdot 3\right) + {6}^{3 x} \left(8 {x}^{3}\right)$

Oct 24, 2015

See the explanation section, below.

Explanation:

Because it involves two commutative operations (multiplication and addition), the product rule can be written in several orders.

I will use the product rule written $\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$ (where the prime indicates a derivative with respect to $x$.

$y = 2 {x}^{4} \cdot {6}^{3 x}$

$u = 2 {x}^{4}$ so $u ' = 8 {x}^{3}$ and

$v = {6}^{3 x}$, so $v ' = {6}^{3 x} \ln 6 \frac{d}{\mathrm{dx}} \left(3 x\right) = 3 \cdot {6}^{3 x} \ln 6$

$y ' = \left(8 {x}^{3}\right) \left({6}^{3 x}\right) + \left(2 {x}^{4}\right) \left(3 \cdot {6}^{3 x} \ln 6\right)$

Rewrite to taste using algebra. I like;

y' = 8x^3 6^(3x) + 6x^4 6^(3x) ln6)

or (better yet)

$y ' = 2 {x}^{3} {6}^{3 x} \left(4 + 3 x \ln 6\right)$