Find y' and y''. y = ln(4x) /x^6

Question

I just need help finding y''

Answer 1

Using the quotient rule: y' = (u'v - uv')v² where:

u = ln(4x)

v = x⁶

u' = 4/4x = 1/x

v' = 6x⁵

v² = x¹²

y' = [(1/x)(x⁶) - (ln(4x))(6x⁵)]/x¹² = (x⁵ - 6x⁵ln(4x))/x¹² = x⁵(1 - 6ln4x))/x¹² = (1 - 6ln(4x))/x⁷

y'' (also by the quotient rule) = (u'v - uv')v² where:

u = 1 - 6ln(4x)

v = x⁷

u' = -24/4x = -6/x

v' = 7x⁶

v² = x¹⁴

y'' = [(-6/x)(x⁷) - (1 - 6ln(4x))(7x⁶)]/x¹⁴

y'' = [-6x⁶ - (7x⁶ - 42x⁶ln(4x))]/x¹⁴

y'' = (-13x6 + 42x⁶ln(4x))/x¹⁴

y'' = x⁶(-13 + 42ln(4x))/x¹⁴

y'' = (-13 + 42ln(4x))/x⁸

y'' = (42ln(4x) - 13)/x⁸

Answer 2

Hi Nicole,

Finding the answers to both of your questions requires using the product rule

When solving problems like this I like to start by rewriting the original equation in a way that makes it clear what f(x) and g(x) are. In this case I'd rewrite the original equation as ln(4x)*(x^-6).

So we can set f(x) = ln(4x) and g(x) =x^-6. Taking the derivatives we get f'(x) = 1/x and g'(x) = -6x^-7. Using the product rule the solution becomes -6*ln(4x)/x^7+1/x^6*(1/x) which can be simplified to:

[1-6*ln(4x)]/x^7

Now we can begin to find the second derivative. Finding the second derivative requires 3 steps. First we must find the derivative of 1/x^7. Second we must complete the product rule to find the derivative of -6*ln(4x)/x^7. Third we must combine the two solutions.

1. 1/x^7 can be rewritten as x^-7. The derivative is then -7x^-8 or -7/x^8.
2. We can rewrite this second piece as ln(4x)*(-6x^-7). This allows us to set our f(x) = ln(4x) and g(x) = -6x^-7. Given these, f'(x) = 1/x and g'(x) = 42x^-8 . Using the product rule our solution becomes 42*ln(4x)/x^8 + -6x^-7*(1/x) which can be reduced to [42*ln(4x) - 6]/x^8
3. Combining the values from the first two steps we get [42*ln(4x) - 6]/x^8 +7/x^8 which reduced to:

[42*ln(4x)-13]/x^8