What is the derivative of 3500(1-1/50 t)^2 ?

The derivative of 3500(1-1/50 t)^2 is -140(1-t/(50))

What is the first derivative of 3500(1-1/50 t)^2 ?

The first derivative of 3500(1-1/50 t)^2 is -140(1-t/(50))

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derivative of

3500 (1 - \frac{1}{50} t)^{2}

- 140 (1 - \frac{t}{50})

Solution steps

\frac{d}{d t} (3500 (1 - \frac{1}{50} t)^{2})

Simplify

3500 (1 - \frac{1}{50} t)^{2} : 3500 (1 - \frac{t}{50})^{2}

= \frac{d}{d t} (3500 (1 - \frac{t}{50})^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3500 \frac{d}{d t} ((1 - \frac{t}{50})^{2})

Apply the chain rule:

2 (1 - \frac{t}{50}) \frac{d}{d t} (1 - \frac{t}{50})

= 2 (1 - \frac{t}{50}) \frac{d}{d t} (1 - \frac{t}{50})

\frac{d}{d t} (1 - \frac{t}{50}) = - \frac{1}{50}

= 3500 \cdot 2 (1 - \frac{t}{50}) (- \frac{1}{50})

Simplify

3500 \cdot 2 (1 - \frac{t}{50}) (- \frac{1}{50}) : - 140 (1 - \frac{t}{50})

= - 140 (1 - \frac{t}{50})

3 t (2 t^{2} - 5)^{4}

\frac{( s - 2 )}{( s - 2 ) ^{2} + 1}

x^{9} d x + y^{9} d y = 0

\frac{1}{s ( s ^{2} + 2 s + 1 )}

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

What is the derivative of 3500(1-1/50 t)^2 ?
The derivative of 3500(1-1/50 t)^2 is -140(1-t/(50))
What is the first derivative of 3500(1-1/50 t)^2 ?
The first derivative of 3500(1-1/50 t)^2 is -140(1-t/(50))