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

The derivative of f(x)=(3x+4)(5-x^2)(4+x^2) is -15x^4-16x^3+9x^2+8x+60

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

The first derivative of f(x)=(3x+4)(5-x^2)(4+x^2) is -15x^4-16x^3+9x^2+8x+60

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derivative of

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

- 15 x^{4} - 16 x^{3} + 9 x^{2} + 8 x + 60

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

3 (5 - x^{2}) (4 + x^{2}) + (- 4 x^{3} + 2 x) (3 x + 4) : - 15 x^{4} - 16 x^{3} + 9 x^{2} + 8 x + 60

= - 15 x^{4} - 16 x^{3} + 9 x^{2} + 8 x + 60

x \to \infty lim (e^{x} + 3)

\int_{0}^{\infty} e^{- \frac{x}{4}} d x

x \to 0 lim ((1 + x)^{\frac{2}{x}})

\frac{d}{d x} (4 e^{3 x} + 12 x e^{3 x})

What is the derivative of f(x)=(3x+4)(5-x^2)(4+x^2) ?
The derivative of f(x)=(3x+4)(5-x^2)(4+x^2) is -15x^4-16x^3+9x^2+8x+60
What is the first derivative of f(x)=(3x+4)(5-x^2)(4+x^2) ?
The first derivative of f(x)=(3x+4)(5-x^2)(4+x^2) is -15x^4-16x^3+9x^2+8x+60