What is the derivative of-1/2 t^4+48t^2 ?

The derivative of-1/2 t^4+48t^2 is -2t^3+96t

What is the first derivative of-1/2 t^4+48t^2 ?

The first derivative of-1/2 t^4+48t^2 is -2t^3+96t

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derivative of

- \frac{1}{2} t^{4} + 48 t^{2}

- 2 t^{3} + 96 t

Solution steps

\frac{d}{d t} (- \frac{1}{2} t^{4} + 48 t^{2})

Simplify

- \frac{1}{2} t^{4} + 48 t^{2} : - \frac{t ^{4}}{2} + 48 t^{2}

= \frac{d}{d t} (- \frac{t ^{4}}{2} + 48 t^{2})

Apply the Sum/Difference Rule:

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

= - \frac{d}{d t} (\frac{t ^{4}}{2}) + \frac{d}{d t} (48 t^{2})

\frac{d}{d t} (\frac{t ^{4}}{2}) = 2 t^{3}

\frac{d}{d t} (48 t^{2}) = 96 t

= - 2 t^{3} + 96 t

f (x) = \frac{1}{x}, at x = 2

\frac{d}{d x} (\frac{d y}{d x}) + 4 y = 2 sin (2 x)

2 x^{^{''}} + 12 x^{^{'}} + 50 x = 0

80 \cdot 2 2^{\frac{t}{20}}

\int \frac{6}{9 + 6 x} d x

What is the derivative of-1/2 t^4+48t^2 ?
The derivative of-1/2 t^4+48t^2 is -2t^3+96t
What is the first derivative of-1/2 t^4+48t^2 ?
The first derivative of-1/2 t^4+48t^2 is -2t^3+96t