# How do you differentiate g(y) =2x^4 * 6^(3x using the product rule?

Aug 16, 2017

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

#### Explanation:

We will need the chain rule along with a handful of standard derivatives:

 {: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (af(x), af'(x), a " constant"), (x^n, nx^(n-1), n " constant"), (a^x, lna \ a^x, ), (f(g(x)), f'(g(x)) \ g'(x),"chain rule" ), (uv, uv'+u'v,"product rule" ) :}

And so if we have:

$g \left(y\right) = 2 {x}^{4} \setminus {6}^{3 x}$

Then differentiating wrt $x$ we have:

$g ' \left(y\right) = \left(2 {x}^{4}\right) \left(\frac{d}{\mathrm{dx}} {6}^{3 x}\right) + \left(\frac{d}{\mathrm{dx}} 2 {x}^{4}\right) \left({6}^{3 x}\right)$

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

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {6}^{3 x} \left\{{x}^{4} \left(\ln 6\right) + 8 {x}^{3}\right\}$