if u differentiate 1/t-4 isnt it -1/t-4? the back of the book says -1/(t-4)^2... which is it please explain!?

Answer 1

Method 1
Let y = 1 / (t - 4) = 1 / u = u^(-1)
dy/du = - u^(-2) = - 1 / u²
u = t - 4
du/dt = 1
dy/dt = (dy/du).(du/dt)
dy/dt = (- 1 / u²).1
dy/dt = - 1 / (t - 4)²
Method 2
f(t) =(t - 4)^(-1)
f `(t) = (-1).(t - 4)^(-2)
f `(t) = -1 / (t - 4)²

Answer 2

The derivative of x*n is always nx^n-1. In this case, x = t-4, and n = -1, so the derivative is -1/(t-4)^2, and the book is correct

Answer 3

Book is right!

Recall that derivative of x^n is n*x^(n - 1)....(1)

So if you take (t - 4) as x, it means that the derivative of 1/x or x^( -1) is required which would be -1/x^2 = - /(t - 4)^2.
Now simple, I hope!

Answer 4

y=1/(t-4)=(t-4)^-1 ---> y'=(t-4)^-2*-1*1= -1/(t-4)^2

y=u^n ---> y'=n*u'*u^(n-1)