How do you differentiate #2x^4 * 6^(3x)#?

1 Answer
Lovecraft
Oct 8, 2015

#d/dx(f(x)) = 6^(3x)*2x^3(ln(6)*3x + 4)#

Explanation:

You can either use the product / chain rule, or implicit differentiation with logarithms. Since you've posted this in Basic Differentiation Rules, I'll be using the chain rule.

Given that

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

We need to find the derivatives of each term to apply the product rule, so, for the first term we have

#dy/dx = 8x^3#

For the second term we have

#y = e^(ln(6)*3x)#

Decomposing it, we have two functions

#y=f(u) = e^u#
#u=g(x) = ln(6)*3x#

The chain rule state that

#dy/dx = dy/(du)*(du)/dx#

Or,

#dy/dx = e^(ln(6)*3x)*ln(6)*3#

So, using the product rule we have

#d/dx(f(x)) = 6^(3x)*ln(6)*6x^4 + 8x^3*6^(3x)#

Or

#d/dx(f(x)) = 6^(3x)*2x^3(ln(6)*3x + 4)#

Doing implicit differentiation with logarithms,

#y = 2x^4 * 6^(3x)#
#ln(y) = ln(2x^4) + ln(6^(3x))#
#ln(y) = ln(2) + 4ln(x) + 3xln(6)#
#y^'/y = 4/x + 3ln(6)#
#y^' = (4(2x^4 * 6^(3x)))/x + 3ln(6)(2x^4 * 6^(3x))#
#y^' = 8x^3*6^(3x) + 3ln(6)(2x^4*6^(3x))#
#y^' = 6^(3x)*2x^3(ln(6)*3x+4)#

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