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

The derivative of f(x)=4x^2+1 is 8x

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

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

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derivative of

f (x) = 4 x^{2} + 1

8 x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 8 x + 0

Simplify

= 8 x

y^{^{'}} + \frac{1}{2} y = 8, y (0) = 2

\frac{\partial}{\partial x} (\frac{( x v ^{2} )}{( x ^{2} + v ^{2} )})

\int \frac{5}{( 5 x + 4 ) ln ( 5 x + 4 )} d x

\int 6 x^{2} - 2 x + 1 d x

n = 1 \sum \infty \frac{n}{n ^{3} + 3}

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