What is the derivative of f(x)= 1/5 x^4-4x^2+1 ?

The derivative of f(x)= 1/5 x^4-4x^2+1 is (4x^3)/5-8x

What is the first derivative of f(x)= 1/5 x^4-4x^2+1 ?

The first derivative of f(x)= 1/5 x^4-4x^2+1 is (4x^3)/5-8x

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derivative of

f (x) = \frac{1}{5} x^{4} - 4 x^{2} + 1

\frac{4 x ^{3}}{5} - 8 x

Solution steps

\frac{d}{d x} (\frac{1}{5} x^{4} - 4 x^{2} + 1)

Simplify

\frac{1}{5} x^{4} - 4 x^{2} + 1 : \frac{x ^{4}}{5} - 4 x^{2} + 1

= \frac{d}{d x} (\frac{x ^{4}}{5} - 4 x^{2} + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{x ^{4}}{5}) - \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (1)

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

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

\frac{d}{d x} (1) = 0

= \frac{4 x ^{3}}{5} - 8 x + 0

Simplify

= \frac{4 x ^{3}}{5} - 8 x

\frac{d}{d x} (\frac{3 x + 2}{2 x + 1})

(x + 1) y^{^{'}} = x + 6

\frac{\partial}{\partial x} (3 x^{4} - sin (5 x) + 4 x)

x \to - 2 lim (x + 2)

(y - x) y^{^{'}} = y - x + 8

What is the derivative of f(x)= 1/5 x^4-4x^2+1 ?
The derivative of f(x)= 1/5 x^4-4x^2+1 is (4x^3)/5-8x
What is the first derivative of f(x)= 1/5 x^4-4x^2+1 ?
The first derivative of f(x)= 1/5 x^4-4x^2+1 is (4x^3)/5-8x