What is the derivative of 6+1/8 (t-4)^2 ?

The derivative of 6+1/8 (t-4)^2 is t/4-1

What is the first derivative of 6+1/8 (t-4)^2 ?

The first derivative of 6+1/8 (t-4)^2 is t/4-1

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derivative of

6 + \frac{1}{8} (t - 4)^{2}

\frac{t}{4} - 1

Solution steps

\frac{d}{d t} (6 + \frac{1}{8} (t - 4)^{2})

Simplify

6 + \frac{1}{8} (t - 4)^{2} : 6 + \frac{( t - 4 ) ^{2}}{8}

= \frac{d}{d t} (6 + \frac{( t - 4 ) ^{2}}{8})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (6) + \frac{d}{d t} (\frac{( t - 4 ) ^{2}}{8})

\frac{d}{d t} (6) = 0

\frac{d}{d t} (\frac{( t - 4 ) ^{2}}{8}) = \frac{t}{4} - 1

= 0 + \frac{t}{4} - 1

Simplify

= \frac{t}{4} - 1

\int π t cos (n t) d t

x \to 0 lim (\frac{2 x}{e ^{x} - 1})

\int \frac{1}{x ^{3} - 216} d x

x = - 2, x = 1, y = 10 x, y = x^{2} - 11

\frac{d}{d x} (\frac{x ^{4} - x ^{3} - 8}{( x - 3 ) ( x + 2 )})

What is the derivative of 6+1/8 (t-4)^2 ?
The derivative of 6+1/8 (t-4)^2 is t/4-1
What is the first derivative of 6+1/8 (t-4)^2 ?
The first derivative of 6+1/8 (t-4)^2 is t/4-1