Please help! Calculus!

y =14^(-8/x)
dy/dx = ?

Let -8/x = u

y = 14^u

The derivative of functions like this: a^b us (a^b) * d/db [b] * ln(a)

y = 14^u

u = -8/x
d/du = 8 / x^2

(14^(-8/x)) * (8 / x^2) * ln(14)

gunna have to do logarithmic differentiating so do

ln(y) = ln( 14 ^ ( -8/x ) ) also rewrite -8/x as -8x^-1
= 14ln(-8x^-1) now do dy/dx and

1/y (dy/dx) y = 14 ( 1 / (-8x^-1) ) + ln(-8x^-1)(0) now simplify ....

1/y (dy/dx)y = -112x
now solve for y

y = (-112x)y
you see 2 ys now the one in the equation simply becomes what you started off with in...

y = 14 ^ (-8 / x)

thus

y = (-122x)( 14 ^ (-8 / x) )

1+1=2 your welcome