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

2 Answers
Oct 24, 2015

Answer:

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

Explanation:

The product rule states that

#d/dx[f(x)*g(x)]=f(x)*g'(x)+g(x)*f'(x)#

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

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

Oct 24, 2015

Answer:

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 #d/dx(uv) = u'v+uv'# (where the prime indicates a derivative with respect to #x#.

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

#u = 2x^4# so #u' = 8x^3# and

#v = 6^(3x)#, so #v' = 6^(3x) ln6 d/dx(3x) = 3*6^(3x) ln6#

#y' = (8x^3)(6^(3x))+(2x^4)(3*6^(3x) ln6)#

Rewrite to taste using algebra. I like;

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

or (better yet)

#y' = 2x^3 6^(3x)(4 + 3x ln6)#