What is the derivative of f(t)=(t^{62}+3t+3)(5t^2+4) ?

The derivative of f(t)=(t^{62}+3t+3)(5t^2+4) is 320t^{63}+248t^{61}+45t^2+30t+12

What is the first derivative of f(t)=(t^{62}+3t+3)(5t^2+4) ?

The first derivative of f(t)=(t^{62}+3t+3)(5t^2+4) is 320t^{63}+248t^{61}+45t^2+30t+12

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derivative of

f (t) = (t^{62} + 3 t + 3) (5 t^{2} + 4)

320 t^{63} + 248 t^{61} + 45 t^{2} + 30 t + 12

Solution steps

\frac{d}{d t} ((t^{62} + 3 t + 3) (5 t^{2} + 4))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{62} + 3 t + 3, g = 5 t^{2} + 4

= \frac{d}{d t} (t^{62} + 3 t + 3) (5 t^{2} + 4) + \frac{d}{d t} (5 t^{2} + 4) (t^{62} + 3 t + 3)

\frac{d}{d t} (t^{62} + 3 t + 3) = 62 t^{61} + 3

\frac{d}{d t} (5 t^{2} + 4) = 10 t

= (62 t^{61} + 3) (5 t^{2} + 4) + 10 t (t^{62} + 3 t + 3)

Simplify

(62 t^{61} + 3) (5 t^{2} + 4) + 10 t (t^{62} + 3 t + 3) : 320 t^{63} + 248 t^{61} + 45 t^{2} + 30 t + 12

= 320 t^{63} + 248 t^{61} + 45 t^{2} + 30 t + 12

\frac{d}{d x} (ln (7 x))

\frac{\partial}{\partial x} ((x y - 1)^{2})

\int \frac{x ^{4} + 4}{x ^{3}} d x

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 2 t}

\int \frac{x ^{2}}{( x ^{2} + a ^{2} ) ^{3}} d x

What is the derivative of f(t)=(t^{62}+3t+3)(5t^2+4) ?
The derivative of f(t)=(t^{62}+3t+3)(5t^2+4) is 320t^{63}+248t^{61}+45t^2+30t+12
What is the first derivative of f(t)=(t^{62}+3t+3)(5t^2+4) ?
The first derivative of f(t)=(t^{62}+3t+3)(5t^2+4) is 320t^{63}+248t^{61}+45t^2+30t+12