What is the derivative of f(x)=8x^4-3x^3+4x ?

The derivative of f(x)=8x^4-3x^3+4x is 32x^3-9x^2+4

What is the first derivative of f(x)=8x^4-3x^3+4x ?

The first derivative of f(x)=8x^4-3x^3+4x is 32x^3-9x^2+4

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derivative of

f (x) = 8 x^{4} - 3 x^{3} + 4 x

32 x^{3} - 9 x^{2} + 4

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (8 x^{4}) = 32 x^{3}

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

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

= 32 x^{3} - 9 x^{2} + 4

\frac{d}{d x} (cos (sin (x^{4})))

\frac{d}{d y} (y^{1 - n})

\int \frac{2 x ^{2}}{4 + x ^{2}} d x

y^{3} - (6 x + 4) + 3 x y^{2} y^{^{'}} = 0

n = 1 \sum \infty \frac{2 ^{n}}{n} (4 x - 8)^{n}

What is the derivative of f(x)=8x^4-3x^3+4x ?
The derivative of f(x)=8x^4-3x^3+4x is 32x^3-9x^2+4
What is the first derivative of f(x)=8x^4-3x^3+4x ?
The first derivative of f(x)=8x^4-3x^3+4x is 32x^3-9x^2+4