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What is the first derivative of r with respect to t (i.e., differentiate r with respect to t)? r = 5/(t2)Note: Use ^ to show exponents in your answer, so for example x2 = x^2. Also, type your equation answer without additional spaces.

User Nerdmaster
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1 Answer

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Answer:

The first derivative of
r(t) = 5\cdot t^(-2) (r(t)=5*t^{-2}) with respect to t is
r'(t) = -10\cdot t^(-3) (r'(t) = -10*t^{-3}).

Explanation:

Let be
r(t) = (5)/(t^(2)), which can be rewritten as
r(t) = 5\cdot t^(-2). The rule of differentiation for a potential function multiplied by a constant is:


(d)/(dt)(c \cdot t^(n)) = n\cdot c \cdot t^(n-1),
\forall \,n\\eq 0

Then,


r'(t) = (-2)\cdot 5\cdot t^(-3)


r'(t) = -10\cdot t^(-3) (r'(t) = -10*t^{-3})

The first derivative of
r(t) = 5\cdot t^(-2) (r(t)=5*t^{-2}) with respect to t is
r'(t) = -10\cdot t^(-3) (r'(t) = -10*t^{-3}).

User Ypakala
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