What is the derivative of f(x)=(x^2+3)^3*4x^2 ?

The derivative of f(x)=(x^2+3)^3*4x^2 is 4(6x^3(x^2+3)^2+2x(x^2+3)^3)

What is the first derivative of f(x)=(x^2+3)^3*4x^2 ?

The first derivative of f(x)=(x^2+3)^3*4x^2 is 4(6x^3(x^2+3)^2+2x(x^2+3)^3)

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derivative of

f (x) = (x^{2} + 3)^{3} \cdot 4 x^{2}

4 (6 x^{3} (x^{2} + 3)^{2} + 2 x (x^{2} + 3)^{3})

Solution steps

\frac{d}{d x} ((x^{2} + 3)^{3} \cdot 4 x^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} ((x^{2} + 3)^{3} x^{2})

Apply the Product Rule:

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

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

= 4 (\frac{d}{d x} ((x^{2} + 3)^{3}) x^{2} + \frac{d}{d x} (x^{2}) (x^{2} + 3)^{3})

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

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

= 4 (6 x (x^{2} + 3)^{2} x^{2} + 2 x (x^{2} + 3)^{3})

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

= 4 (6 x^{3} (x^{2} + 3)^{2} + 2 x (x^{2} + 3)^{3})

x \to 0 lim (\frac{( ln ( x ) ) ^{2}}{x})

y^{^{'}} = (x + y)^{2}

\frac{1 + csc ( x ^{3} )}{1 - cot ( x ^{3} )}

\frac{d}{d x} (24)

\frac{\partial}{\partial v} (3 u)

What is the derivative of f(x)=(x^2+3)^3*4x^2 ?
The derivative of f(x)=(x^2+3)^3*4x^2 is 4(6x^3(x^2+3)^2+2x(x^2+3)^3)
What is the first derivative of f(x)=(x^2+3)^3*4x^2 ?
The first derivative of f(x)=(x^2+3)^3*4x^2 is 4(6x^3(x^2+3)^2+2x(x^2+3)^3)