What is the derivative of h(t)=(t^4-1)^3(t^3+1)^4 ?

The derivative of h(t)=(t^4-1)^3(t^3+1)^4 is 12t^3(t^4-1)^2(t^3+1)^4+12t^2(t^3+1)^3(t^4-1)^3

What is the first derivative of h(t)=(t^4-1)^3(t^3+1)^4 ?

The first derivative of h(t)=(t^4-1)^3(t^3+1)^4 is 12t^3(t^4-1)^2(t^3+1)^4+12t^2(t^3+1)^3(t^4-1)^3

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derivative of

h (t) = (t^{4} - 1)^{3} (t^{3} + 1)^{4}

12 t^{3} (t^{4} - 1)^{2} (t^{3} + 1)^{4} + 12 t^{2} (t^{3} + 1)^{3} (t^{4} - 1)^{3}

Solution steps

\frac{d}{d t} ((t^{4} - 1)^{3} (t^{3} + 1)^{4})

Apply the Product Rule:

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

f = (t^{4} - 1)^{3}, g = (t^{3} + 1)^{4}

= \frac{d}{d t} ((t^{4} - 1)^{3}) (t^{3} + 1)^{4} + \frac{d}{d t} ((t^{3} + 1)^{4}) (t^{4} - 1)^{3}

\frac{d}{d t} ((t^{4} - 1)^{3}) = 12 t^{3} (t^{4} - 1)^{2}

\frac{d}{d t} ((t^{3} + 1)^{4}) = 12 t^{2} (t^{3} + 1)^{3}

= 12 t^{3} (t^{4} - 1)^{2} (t^{3} + 1)^{4} + 12 t^{2} (t^{3} + 1)^{3} (t^{4} - 1)^{3}

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\frac{d}{d x} (x e^{- \frac{1}{2} x})

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x \to 1 lim (\frac{3 sin ( π x ) - sin ( 3 π x )}{x ^{3}})

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What is the derivative of h(t)=(t^4-1)^3(t^3+1)^4 ?
The derivative of h(t)=(t^4-1)^3(t^3+1)^4 is 12t^3(t^4-1)^2(t^3+1)^4+12t^2(t^3+1)^3(t^4-1)^3
What is the first derivative of h(t)=(t^4-1)^3(t^3+1)^4 ?
The first derivative of h(t)=(t^4-1)^3(t^3+1)^4 is 12t^3(t^4-1)^2(t^3+1)^4+12t^2(t^3+1)^3(t^4-1)^3