What is the derivative of 3x^2(4x^2-3^5) ?

The derivative of 3x^2(4x^2-3^5) is 3(2x(4x^2-3)^5+40x^3(4x^2-3)^4)

What is the first derivative of 3x^2(4x^2-3^5) ?

The first derivative of 3x^2(4x^2-3^5) is 3(2x(4x^2-3)^5+40x^3(4x^2-3)^4)

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\frac{d}{d x} (3 x^{2} (4 x^{2} - 3)^{5})

3 (2 x (4 x^{2} - 3)^{5} + 40 x^{3} (4 x^{2} - 3)^{4})

Solution steps

\frac{d}{d x} (3 x^{2} (4 x^{2} - 3)^{5})

Take the constant out:

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

= 3 \frac{d}{d x} (x^{2} (4 x^{2} - 3)^{5})

Apply the Product Rule:

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

f = x^{2}, g = (4 x^{2} - 3)^{5}

= 3 (\frac{d}{d x} (x^{2}) (4 x^{2} - 3)^{5} + \frac{d}{d x} ((4 x^{2} - 3)^{5}) x^{2})

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

\frac{d}{d x} ((4 x^{2} - 3)^{5}) = 40 x (4 x^{2} - 3)^{4}

= 3 (2 x (4 x^{2} - 3)^{5} + 40 x (4 x^{2} - 3)^{4} x^{2})

40 x (4 x^{2} - 3)^{4} x^{2} = 40 x^{3} (4 x^{2} - 3)^{4}

= 3 (2 x (4 x^{2} - 3)^{5} + 40 x^{3} (4 x^{2} - 3)^{4})

x \to 2 lim (5 x^{2} - 12)

\frac{\partial}{\partial x} (4 x^{2} + y^{2} - 5 y)

\int ln (x + (x^{2} + 1)) d x

\int_{- 1}^{2} 4 y^{3} x d y

\int 2 sin (x) cos (cos (x)) d x

What is the derivative of 3x^2(4x^2-3^5) ?
The derivative of 3x^2(4x^2-3^5) is 3(2x(4x^2-3)^5+40x^3(4x^2-3)^4)
What is the first derivative of 3x^2(4x^2-3^5) ?
The first derivative of 3x^2(4x^2-3^5) is 3(2x(4x^2-3)^5+40x^3(4x^2-3)^4)