What is the d/(dt)(5t^4+4t^3+2t^2+5+4) ?

The d/(dt)(5t^4+4t^3+2t^2+5+4) is 20t^3+12t^2+4t

What is the first d/(dt)(5t^4+4t^3+2t^2+5+4) ?

The first d/(dt)(5t^4+4t^3+2t^2+5+4) is 20t^3+12t^2+4t

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\frac{d}{d t} (5 t^{4} + 4 t^{3} + 2 t^{2} + 5 + 4)

20 t^{3} + 12 t^{2} + 4 t

Solution steps

\frac{d}{d t} (5 t^{4} + 4 t^{3} + 2 t^{2} + 5 + 4)

Simplify

5 t^{4} + 4 t^{3} + 2 t^{2} + 5 + 4 : 5 t^{4} + 4 t^{3} + 2 t^{2} + 9

= \frac{d}{d t} (5 t^{4} + 4 t^{3} + 2 t^{2} + 9)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (5 t^{4}) + \frac{d}{d t} (4 t^{3}) + \frac{d}{d t} (2 t^{2}) + \frac{d}{d t} (9)

\frac{d}{d t} (5 t^{4}) = 20 t^{3}

\frac{d}{d t} (4 t^{3}) = 12 t^{2}

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

\frac{d}{d t} (9) = 0

= 20 t^{3} + 12 t^{2} + 4 t + 0

Simplify

= 20 t^{3} + 12 t^{2} + 4 t

\frac{d}{d r} (\frac{g \cdot ( m _{1} \cdot m _{2} )}{( r ^{2} )})

5 x^{2} - 5 x + 3

\frac{d}{d a} (2 sin^{2} (a) cos^{2} (a))

f (x) = x^{3} sin^{2} (x)

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

What is the d/(dt)(5t^4+4t^3+2t^2+5+4) ?
The d/(dt)(5t^4+4t^3+2t^2+5+4) is 20t^3+12t^2+4t
What is the first d/(dt)(5t^4+4t^3+2t^2+5+4) ?
The first d/(dt)(5t^4+4t^3+2t^2+5+4) is 20t^3+12t^2+4t