What is the derivative of h(t)=(t^4-1)^9(t^3+1)^6 ?

The derivative of h(t)=(t^4-1)^9(t^3+1)^6 is 36t^3(t^4-1)^8(t^3+1)^6+18t^2(t^3+1)^5(t^4-1)^9

What is the first derivative of h(t)=(t^4-1)^9(t^3+1)^6 ?

The first derivative of h(t)=(t^4-1)^9(t^3+1)^6 is 36t^3(t^4-1)^8(t^3+1)^6+18t^2(t^3+1)^5(t^4-1)^9

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derivative of

h (t) = (t^{4} - 1)^{9} (t^{3} + 1)^{6}

36 t^{3} (t^{4} - 1)^{8} (t^{3} + 1)^{6} + 18 t^{2} (t^{3} + 1)^{5} (t^{4} - 1)^{9}

Solution steps

\frac{d}{d t} ((t^{4} - 1)^{9} (t^{3} + 1)^{6})

Apply the Product Rule:

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

f = (t^{4} - 1)^{9}, g = (t^{3} + 1)^{6}

= \frac{d}{d t} ((t^{4} - 1)^{9}) (t^{3} + 1)^{6} + \frac{d}{d t} ((t^{3} + 1)^{6}) (t^{4} - 1)^{9}

\frac{d}{d t} ((t^{4} - 1)^{9}) = 36 t^{3} (t^{4} - 1)^{8}

\frac{d}{d t} ((t^{3} + 1)^{6}) = 18 t^{2} (t^{3} + 1)^{5}

= 36 t^{3} (t^{4} - 1)^{8} (t^{3} + 1)^{6} + 18 t^{2} (t^{3} + 1)^{5} (t^{4} - 1)^{9}

n = 1 \sum \infty 2 + 7 n

\frac{\partial}{\partial x} (- \frac{1}{2} ln (\frac{x}{a}))

f (x) = x (x - 8), at x = 16

\int_{0}^{3} 6 y - 2 y^{2} d y

ln (2 + x)

What is the derivative of h(t)=(t^4-1)^9(t^3+1)^6 ?
The derivative of h(t)=(t^4-1)^9(t^3+1)^6 is 36t^3(t^4-1)^8(t^3+1)^6+18t^2(t^3+1)^5(t^4-1)^9
What is the first derivative of h(t)=(t^4-1)^9(t^3+1)^6 ?
The first derivative of h(t)=(t^4-1)^9(t^3+1)^6 is 36t^3(t^4-1)^8(t^3+1)^6+18t^2(t^3+1)^5(t^4-1)^9