y = 8^((x^2)-3x+12) , find y' when x=2

The answer I end up with is: y'=(2(2)-3)(e^((4-6+12)*ln8) y'=e^(10*ln8)
The deivative of a^f(x) is a^f'(x) ln (a), where a is a consatnt. y=8^((x^2)-3x+12) y' = 8^((x^2)-3x+12) (2x-3) ln (8) y'= 8(2x-3) 8^((x^2)-3x+12) when x=2 y'=8(1)8^(4-6+12) =8 8^10
=8^11
lny=(x^2-3x+12)ln8

1/y(dy/dx) = (2x - 3)ln8

then you multiply by y, remembring that you had y=.. at the start an substituting that in

dy/dx = 8^(x^2-3x+12)*(2x-3)ln8

so when x=2 dy/dx = 8^(4 - 6 +12) * ln8 dy/dx = 8^10*ln8 = 2232783354 to a whopping 10 s.f. but they usually just want the answer in terms of ln ln(a / b) does not equal ln(a) / ln(b), from the property of logs, split the log into two separate logs through subtraction:

y = ln(x + 2) - ln(x - 1)

Now take the derivative of each term:

dy/dx = [1/(x + 2) * 1] - [1/(x - 1) * 1]

dy/dx = 1 / (x + 2) - 1 / (x - 1)

Optionally, you can make the denominators common and combine them into a single fraction:

dy/dx = -3 / (x - 1)(x + 2) ln [base m] x = ln(x) / ln(m)
This is how you convert log of any base to log of base e.

log[base5] (x + 2) = ln(x+2) /ln(5) log[base7] { ln(x+2) /ln(5) } = ln (ln(x+2) /ln(5)) / ln(7) log[base3] { ln (ln(x+2) /ln(5)) / ln(7) } = ln ( ln (ln(x+2) /ln(5)) / ln(7) ) / ln(3)

f(x) = [1/ln(3)] ln ( ln (ln(x+2) /ln(5)) / ln(7) ) let k = [1/ln(3)]
Let ln (x+2) /ln(5) = u f(x) = k ln ( ln (u) /ln(7) )

f'(x) = k [1 / ( ln (u) /ln(7) ) ] d/dx[ ( ln (u) /ln(7) ) ---(1) d/dx [ ln (u) / ln(7) ] = 1/ln(7) (1/u) du/dx ---(2) du/dx = 1/ln(5)(x+2)
(2) becomes 1/ln(7) (1/u) 1/ln(5)(x+2)
= [1/ln(7)] [ln(5) /ln(x+2) ][ 1/ln(5)(x+2)]
= 1/ ln(7) ln(x+2) (x+2)
(1) becomes k [1 / ( ln (u) /ln(7) ) [1/ ln(7) ln(x+2) (x+2)]
= k [ln(7) /ln(u) ][1/ln(7) ln(x+2)(x+2)]
= k / ln(u) ln(x+2) (x+2)

= k / ln ( ln (x+2) /ln(5)) ln(x+2) (x+2)
= 1 / ln(3)