What is the derivative of f(t)=8t-9t^2 ?

The derivative of f(t)=8t-9t^2 is 8-18t

What is the first derivative of f(t)=8t-9t^2 ?

The first derivative of f(t)=8t-9t^2 is 8-18t

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derivative of

f (t) = 8 t - 9 t^{2}

8 - 18 t

Solution steps

\frac{d}{d t} (8 t - 9 t^{2})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (8 t) - \frac{d}{d t} (9 t^{2})

\frac{d}{d t} (8 t) = 8

\frac{d}{d t} (9 t^{2}) = 18 t

= 8 - 18 t

\frac{\partial}{\partial x} (e^{x y} + 2 x - 5 y)

y^{^{''}} + 8 y^{^{'}} + 7 y = 0, y (0) = 6, y^{^{'}} (0) = - 1

\frac{d ^{2}}{d x ^{2}} (x^{4} - 6 x^{3})

\frac{\partial}{\partial x} (\frac{7 x}{x ^{2} + y ^{2}})

sin (6 (t - \frac{5 π}{6}))

What is the derivative of f(t)=8t-9t^2 ?
The derivative of f(t)=8t-9t^2 is 8-18t
What is the first derivative of f(t)=8t-9t^2 ?
The first derivative of f(t)=8t-9t^2 is 8-18t