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

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

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

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

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derivative of

f (x) = \frac{x ^{4} + c _{1}}{4 x ^{2}}

\frac{x ^{4} - c _{1}}{2 x ^{3}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

= \frac{1}{4} \cdot \frac{\frac{d}{d x} ( x ^{4} + c _{1} ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( x ^{4} + c _{1} )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (x^{4} + c_{1}) = 4 x^{3}

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

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

Simplify

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

= \frac{x ^{4} - c _{1}}{2 x ^{3}}

\int 55 - 6 x - x^{2} d x

n \to \infty lim (28 n + 4)

\frac{a x + b}{c x + d}

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

\frac{d}{d x} (2 x e^{2 x} + 2 x^{2} e^{2 x})

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