Question

Find the derivative using first principles? : `G(t) = (1-5t)/(4+t)`

Answer 1

`G'(t) = -21/(4+t)^2`

Explanation:

We have:

`G(t) = (1-5t)/(4+t)`

Then the limit definition of the derivative tells us that;

`G'(t) = lim_(h rarr 0) ( f(t+h)-f(t) ) / h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( (1-5(t+h))/(4+(t+h)) - (1-5t)/(4+t) ) / h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( ( (1-5t-5h)(4+t) - (1-5t)(4+t+h) ) / ((4+t+h)(4+t)) ) / h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( (4+t-20t-5t^2-20h-5ht) - (4-20t+t-5t^2+h-5h) ) / (h(4+t+h)(4+t))`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 4+t-20t-5t^2-20h-5ht -4+20t-t+5t^2-h+5ht ) / (h(4+t+h)(4+t))`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( -21h ) / (h(4+t+h)(4+t))`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( -21 ) / ((4+t+h)(4+t))`

`\ \ \ \ \ \ \ \ \ = -21 / (4+t)^2`