What is the derivative of f(x)=(x^2+3)^3(x^3+3)^2 ?

The derivative of f(x)=(x^2+3)^3(x^3+3)^2 is 6x(x^2+3)^2(x^3+3)^2+6x^2(x^3+3)(x^2+3)^3

What is the first derivative of f(x)=(x^2+3)^3(x^3+3)^2 ?

The first derivative of f(x)=(x^2+3)^3(x^3+3)^2 is 6x(x^2+3)^2(x^3+3)^2+6x^2(x^3+3)(x^2+3)^3

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derivative of

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

6 x (x^{2} + 3)^{2} (x^{3} + 3)^{2} + 6 x^{2} (x^{3} + 3) (x^{2} + 3)^{3}

Solution steps

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

Apply the Product Rule:

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

f = (x^{2} + 3)^{3}, g = (x^{3} + 3)^{2}

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

\frac{d}{d x} ((x^{2} + 3)^{3}) = 6 x (x^{2} + 3)^{2}

\frac{d}{d x} ((x^{3} + 3)^{2}) = 6 x^{2} (x^{3} + 3)

= 6 x (x^{2} + 3)^{2} (x^{3} + 3)^{2} + 6 x^{2} (x^{3} + 3) (x^{2} + 3)^{3}

\frac{d}{d x} (5 x^{2} cos (7 x))

f^{^{''}} (x) = x^{3} - 4 x + 1, f^{^{'}} (0) = 2, f (1) = - 2

x \to 2 lim (x^{2} - 5 x + 6 x^{2} - 4)

x \to 7 lim (\frac{x + 7}{x - 7})

\int cos (8 x) cos (3 x) d x

What is the derivative of f(x)=(x^2+3)^3(x^3+3)^2 ?
The derivative of f(x)=(x^2+3)^3(x^3+3)^2 is 6x(x^2+3)^2(x^3+3)^2+6x^2(x^3+3)(x^2+3)^3
What is the first derivative of f(x)=(x^2+3)^3(x^3+3)^2 ?
The first derivative of f(x)=(x^2+3)^3(x^3+3)^2 is 6x(x^2+3)^2(x^3+3)^2+6x^2(x^3+3)(x^2+3)^3