Whats the derivative of s(t) = 0.5t^2? i wanna check sumthing..

S(t) follows the model: f(x) 
The derivative function, also known as the function gradient is modeled by f'(x). In your case, the derivative of s(t) is s'(t). We use the derivative, to calculate the gradient of a chord/slope. 
s(t) = 0.5t² If we differentiate this function, we carry the exponent and multiply it with the coefficient, and take ONE from the exponent. 
NOTE: if the exponent was 4, after its differentiated, it will become 3. In your case, the exponent is 2, so after differentiation, it will become a 1. Knowing the rule, anything to the power of 1, will remain constant/unchanged. Say, we want to differentiate, x². After differentiation it will become 2x. Rather than 2x^1.

Also, include a dash [s'(t)] to indicate that it's already differentiated.
Thus,
s'(t) = 2×0.5t
∴     = 1t
Which can be wrote as t.
You're answer is;
s'(t) = t

You can differentiate a function TWICE. So if it's needed, you should do so. It will help you find the gradient of the chord. 

I hope you followed through.

S(t) = 0.5t^2

s'(t) = 0.5*2*t^(2 - 1)            using formula for derivative.

s'(t) = 1*t^(1)

 s'(t) = t^(1)

 s'(t) = t

Derivative = t

300b(x) is graphed by the functio