Question

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

Answer 1

`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)`