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

1 Answer
Aug 16, 2017

Answer:

# g'(y) = 6^(3x) {x^4(ln6 ) + 8x^3 } #

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(y) = 2x^4 \ 6^(3x) #

Then differentiating wrt #x# we have:

# g'(y) = (2x^4)(d/dx 6^(3x)) + (d/dx 2x^4)( 6^(3x) ) #

# \ \ \ \ \ \ \ \ = (2x^4)(ln6 \ 6^(3x) \ d/dx(3x)) + (8x^3)( 6^(3x) ) #

# \ \ \ \ \ \ \ \ = 2x^4(ln6 \ 6^(3x) \ 3) + 8x^3 \ 6^(3x) #

# \ \ \ \ \ \ \ \ = 6^(3x) {x^4(ln6 ) + 8x^3 } #