What is the derivative of f(t)=(1+sqrt(t))(6t^2-9) ?

The derivative of f(t)=(1+sqrt(t))(6t^2-9) is (30t^2+24tsqrt(t)-9)/(2sqrt(t))

What is the first derivative of f(t)=(1+sqrt(t))(6t^2-9) ?

The first derivative of f(t)=(1+sqrt(t))(6t^2-9) is (30t^2+24tsqrt(t)-9)/(2sqrt(t))

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derivative of

f (t) = (1 + t) (6 t^{2} - 9)

\frac{30 t ^{2} + 24 t t - 9}{2 t}

Solution steps

\frac{d}{d t} ((1 + t) (6 t^{2} - 9))

Apply the Product Rule:

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

f = 1 + t, g = 6 t^{2} - 9

= \frac{d}{d t} (1 + t) (6 t^{2} - 9) + \frac{d}{d t} (6 t^{2} - 9) (1 + t)

\frac{d}{d t} (1 + t) = \frac{1}{2 t}

\frac{d}{d t} (6 t^{2} - 9) = 12 t

= \frac{1}{2 t} (6 t^{2} - 9) + 12 t (1 + t)

Simplify

\frac{1}{2 t} (6 t^{2} - 9) + 12 t (1 + t) : \frac{30 t ^{2} + 24 t t - 9}{2 t}

= \frac{30 t ^{2} + 24 t t - 9}{2 t}

\frac{\partial}{\partial x} (e^{9 z} - x yz)

\int 3 x cot (x^{2}) d x

\frac{d}{d θ} (a sec^{2} (\frac{θ}{2}))

x \to - 1 lim (3 x^{2} - 7 x - 4)

\int (4 x^{3} - 2)^{2} d x

What is the derivative of f(t)=(1+sqrt(t))(6t^2-9) ?
The derivative of f(t)=(1+sqrt(t))(6t^2-9) is (30t^2+24tsqrt(t)-9)/(2sqrt(t))
What is the first derivative of f(t)=(1+sqrt(t))(6t^2-9) ?
The first derivative of f(t)=(1+sqrt(t))(6t^2-9) is (30t^2+24tsqrt(t)-9)/(2sqrt(t))