What is derivative of f(x)=-16t^2-8t+64 ?

The answer to derivative of f(x)=-16t^2-8t+64 is (dt)/(dx)=0

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derivative of f(x)=-16t^2-8t+64

derivative of

f (x) = - 16 t^{2} - 8 t + 64

\frac{d t}{d x} = 0

Solution steps

f (x) = - 16 t^{2} - 8 t + 64

Treat

t

t (x)

Differentiate both sides:

0 = - 32 t \frac{d t}{d x} - 8 \frac{d t}{d x}

Isolate

\frac{d t}{d x} : \frac{d t}{d x} = 0

\frac{d t}{d x} = 0

(x^{2} - 3) (x^{3} - 2 x)

\int_{0}^{\frac{π}{4}} (cos (2 x)) d x

y^{^{''}} + 2 y^{^{'}} + 5 y = - 2 t + 4 t^{2}

\frac{d}{d x} (\frac{3}{x ^{3}} - \frac{1}{x ^{4}})

\frac{d}{d x} (\frac{( x ^{2} - 3 x )}{( 2 x + 1 )})

What is derivative of f(x)=-16t^2-8t+64 ?
The answer to derivative of f(x)=-16t^2-8t+64 is (dt)/(dx)=0