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

The derivative of f(x)=4x^3(2x^2-3x^3) is 4(-18x^5+10x^4)

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

The first derivative of f(x)=4x^3(2x^2-3x^3) is 4(-18x^5+10x^4)

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derivative of

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

4 (- 18 x^{5} + 10 x^{4})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{3}, g = 2 x^{2} - 3 x^{3}

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

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

\frac{d}{d x} (2 x^{2} - 3 x^{3}) = 4 x - 9 x^{2}

= 4 (3 x^{2} (2 x^{2} - 3 x^{3}) + (4 x - 9 x^{2}) x^{3})

Expand

3 x^{2} (2 x^{2} - 3 x^{3}) + (4 x - 9 x^{2}) x^{3} : - 18 x^{5} + 10 x^{4}

= 4 (- 18 x^{5} + 10 x^{4})

\int sec^{2} (x) tan^{8} (x) d x

(3 x + 4)^{^{'}}

\int \frac{1}{x ^{4}} \cdot x^{5} e^{x} d x

\int \frac{1}{( 2 - x ) ^{2} 1 - x} d x

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

What is the derivative of f(x)=4x^3(2x^2-3x^3) ?
The derivative of f(x)=4x^3(2x^2-3x^3) is 4(-18x^5+10x^4)
What is the first derivative of f(x)=4x^3(2x^2-3x^3) ?
The first derivative of f(x)=4x^3(2x^2-3x^3) is 4(-18x^5+10x^4)