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

The derivative of 3x^3(x^2+4^4) is 3(3x^2(x^2+4)^4+8x^4(x^2+4)^3)

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

The first derivative of 3x^3(x^2+4^4) is 3(3x^2(x^2+4)^4+8x^4(x^2+4)^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

3 (3 x^{2} (x^{2} + 4)^{4} + 8 x^{4} (x^{2} + 4)^{3})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

\frac{d}{d x} ((x^{2} + 4)^{4}) = 8 x (x^{2} + 4)^{3}

= 3 (3 x^{2} (x^{2} + 4)^{4} + 8 x (x^{2} + 4)^{3} x^{3})

8 x (x^{2} + 4)^{3} x^{3} = 8 x^{4} (x^{2} + 4)^{3}

= 3 (3 x^{2} (x^{2} + 4)^{4} + 8 x^{4} (x^{2} + 4)^{3})

\int 9 - A^{2} d A

x = y^{2} - 2 y, x - y - 4 = 0

h^{^{'}}

\int cos^{3} (θ) d θ

4 y^{^{''}} + 4 y^{^{'}} + y = 0

What is the derivative of 3x^3(x^2+4^4) ?
The derivative of 3x^3(x^2+4^4) is 3(3x^2(x^2+4)^4+8x^4(x^2+4)^3)
What is the first derivative of 3x^3(x^2+4^4) ?
The first derivative of 3x^3(x^2+4^4) is 3(3x^2(x^2+4)^4+8x^4(x^2+4)^3)