What is the derivative of f(x)=(x^3+1)/(sqrt(x^2+1)) ?

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

What is the first derivative of f(x)=(x^3+1)/(sqrt(x^2+1)) ?

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

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derivative of

f (x) = \frac{x ^{3} + 1}{x ^{2} + 1}

\frac{2 x ^{4} + 3 x ^{2} - x}{( x ^{2} + 1 ) x ^{2} + 1}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

= \frac{3 x ^{2} x ^{2} + 1 - \frac{x}{x ^{2} + 1} ( x ^{3} + 1 )}{( x ^{2} + 1 ) ^{2}}

Simplify

\frac{3 x ^{2} x ^{2} + 1 - \frac{x}{x ^{2} + 1} ( x ^{3} + 1 )}{( x ^{2} + 1 ) ^{2}} : \frac{2 x ^{4} + 3 x ^{2} - x}{( x ^{2} + 1 ) x ^{2} + 1}

= \frac{2 x ^{4} + 3 x ^{2} - x}{( x ^{2} + 1 ) x ^{2} + 1}

\int tan (u) d u

\frac{d y}{d x} x y = 4

x \to 2 lim (\frac{x ^{3} - 8}{x - 4})

x \to 6 lim (\frac{4}{x - 6})

\frac{d ^{2}}{d x ^{2}} (tan (2 x))

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