# Find f '(t) using the definition of derivative. `f(t) = (2t + 1)/(t+3)` My algebra is not coming out like in the back of the book.

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`f(t)=(2t+1)/(t+3)`

The derivative of this function with respect to t can determined using the definition of derivative which is:

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

To apply this formula, evaluate the given function at t=t+h to get f(t+h).

`f(t)=(2t+1)/(t+3)`

`f(t+h)=(2(t+h)+1)/((t+h)+3)`

`f(t+h)=(2t+2h+1)/(t+h+3)`

Then, plug-in this to the formula of above.

`f'(t)=lim_(h->0)((2t+2h+1)/(t+h+3)-(2t+1)/(t+3))/h`

Before taking the limit, simplify the complex fraction. To do so, multiply the top and bottom by the LCD which is (t+h+3)(t+3).

`=((2t+2h+1)/(t+h+3)-(2t+1)/(t+3))/h*((t+h+3)(t+3))/((t+h+3)(t+3))`

`=(((2t+2h+1)/(t+h+3)-(2t+1)/(t+3))(t+h+3)(t+3))/(h(t+h+3)(t+3))`

`=((2t+2h+1)(t+3)-(2t+1)(t+h+3))/(h(t+h+3)(t+3))`

To simplify the numerator, apply distributive property. And combine like terms.

`=(2t^2+2ht+t+6t+6h+3-(2t^2+2ht+6t+t+h+3))/(h(t+h+3)(t+3))`

`=(2t^2+2ht+7t+6h+3-(2t^2+2ht+7t+h+3))/(h+(t+h+3)(t+3))`

`=(2t^2+2ht+7t+6h+3-2t^2-2ht-7t-h-3)/(h(t+h+3)(t+3))`

`=(5h)/(h(t+h+3)(t+3))`

Then, cancel common factor which is h.

`=5/((t+h+3)(t+3))`

So,

`f'(t)=lim_(h->0)5/((t+h+3)(t+3))`

Now that it is in simplified form, proceed to take the limit as h approaches zero. To do so, set h=0.

`f'(t)=5/((t+0+3)(t+3))=5/((t+3)(t+3))`

`f'(t)=5/(t+3)^2`

**Hence, the derivative of the given function with respect to t is `f'(t)=5/(t+3)^2` .**

`f(t)=(2t+1)/(t+3)`

`=(2t+6-5)/(t+3)`

`=2(t+3)/(t+3)-5/(t+3)`

`f(t)=2-5/(t+3)`

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

`f(t+h)-f(t)=2-5/(t+h+3)-(2-5/(t+3))`

`=5/(t+3)-5/(t+h+3)`

`=5(1/(t+3)-1/(t+h+3))`

`=(5(t+h+3-t-3))/((t+3)(t+h+3))`

`=(5h)/((t+3)(t+h+3))`

`f'(t)=lim_(h->0)(5h)/(h(t+3)(t+h+3))`

`=lim_(h->0)5/((t+3)(t+h+3))`

`=5/((t+3)(t+3))`

`so`

` f'(t)=5/(t+3)^2`

ya mean `lim_(h->0) (f(a+h)-f(a))/h`

`f(t)=(2t+1)/(t+3)=(2t+6-5)/(t+3)=2-5/(t+3)`

`f'(t)=(-5)(-1)/(t+3)^2=5/(t+3)^2`