how do you differentiate 8/t^2?

Answer 1

Hi Dear Ned ;

{ if f(x) = u / v so
f '(x) = [( u ' * v) - ( u * v ' ) ] / v ^2 } ...

if f(x) = 8/t^2
u = 8 , u ' = 0
v = t^2 , v ' = 2t

f '(x) = [ ( 0 * t^2 ) - ( 2t * 8 ) ] / (t ^2 )^2 =
f '(x) = - 16t / t ^ 4 =
f '(x) = - 16 ( t ^ ( 1 - 4))
f '(x) = - 16 t ^ -3
f '(x) = - 16 / ( t ^3 )

Good Luck Dear ♣

Answer 2

Okay, to try and tackle this problem, it might be easier to write the expression as 8t^(-2). Then you can just differentiate this with respect to t, using the fact that if you have something of the form x^n that you want to differentiate, then the derivative is going to be nx^(n-1). Thus, when you differentiate 8t^(-2), you get:

8x(-2)t^(-3)
That is, -16t^(-3), or to put it back in the original form:
-16/(t^3).

Hope that this is helpful!

Answer 3

One of the key first steps to differentiate is to get the expression into index form.

8/t^2 = 8t^-2 that is 8 times t to the power of index of -2

The off you go

dy/dt = -2 x 8 x t^-2-1 = -16t^-3

It is usual to write the answer in the form of the original expression.

Hence = -16/t^3

Answer 4

8/t^2 is basically 8*t^(-2), so the answer by normal simple differentiation is 8*(-2)*t^(-3) which is just equal to -16/t^3.

Answer 5

8/t^2 ===> 16/t

Answer 6

As has already been answered, it is -16/t^3.

If you write 8t^-2 then you can see from normal differentiation rules that the derivative is
-16t^-3 = -16/t^3

Answer 7

The t^2 on the bottom is equivalent to t^-2. You then differentiate 8t^-2 which is -16t^-3 (-16/t^3).

Answer 8

8/t^2 is 8

Answer 9

8/t = 5x - z+sub½2/2=x+y

Answer 10

-16/t^3