What is the derivative of f(x)=x^4-20x^2+64 ?

The derivative of f(x)=x^4-20x^2+64 is 4x^3-40x

What is the first derivative of f(x)=x^4-20x^2+64 ?

The first derivative of f(x)=x^4-20x^2+64 is 4x^3-40x

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derivative of

f (x) = x^{4} - 20 x^{2} + 64

4 x^{3} - 40 x

Solution steps

\frac{d}{d x} (x^{4} - 20 x^{2} + 64)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) - \frac{d}{d x} (20 x^{2}) + \frac{d}{d x} (64)

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

\frac{d}{d x} (20 x^{2}) = 40 x

\frac{d}{d x} (64) = 0

= 4 x^{3} - 40 x + 0

Simplify

= 4 x^{3} - 40 x

\frac{d}{d x} (ln (\frac{sin ( x )}{cos ( x )}))

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

\int \frac{sin ( x )}{e ^{2 x}} d x

y^{^{'}} = \frac{x - 1}{y - 1}

\frac{d}{d y} (- \frac{x}{y ^{2}})

What is the derivative of f(x)=x^4-20x^2+64 ?
The derivative of f(x)=x^4-20x^2+64 is 4x^3-40x
What is the first derivative of f(x)=x^4-20x^2+64 ?
The first derivative of f(x)=x^4-20x^2+64 is 4x^3-40x