What is the derivative of ((sqrt(5x^4+x^2))/2) ?

The derivative of ((sqrt(5x^4+x^2))/2) is (10x^2+1)/(2sqrt(5x^2+1))

What is the first derivative of ((sqrt(5x^4+x^2))/2) ?

The first derivative of ((sqrt(5x^4+x^2))/2) is (10x^2+1)/(2sqrt(5x^2+1))

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\frac{d}{d x} (\frac{( 5 x ^{4} + x ^{2} )}{2})

\frac{10 x ^{2} + 1}{2 5 x ^{2} + 1}

Solution steps

\frac{d}{d x} (\frac{( 5 x ^{4} + x ^{2} )}{2})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (5 x^{4} + x^{2})

Apply the chain rule:

\frac{1}{2 5 x ^{4} + x ^{2}} \frac{d}{d x} (5 x^{4} + x^{2})

= \frac{1}{2 5 x ^{4} + x ^{2}} \frac{d}{d x} (5 x^{4} + x^{2})

\frac{d}{d x} (5 x^{4} + x^{2}) = 20 x^{3} + 2 x

= \frac{1}{2} \cdot \frac{1}{2 5 x ^{4} + x ^{2}} (20 x^{3} + 2 x)

Simplify

\frac{1}{2} \cdot \frac{1}{2 5 x ^{4} + x ^{2}} (20 x^{3} + 2 x) : \frac{10 x ^{2} + 1}{2 5 x ^{2} + 1}

= \frac{10 x ^{2} + 1}{2 5 x ^{2} + 1}

(1 + x^{2}) (1 + y^{2}) d x - x y d y = 0

\int (6 sin (3 x) - cos (2 x)) d x

\int x - 4 d x

x \to 0 + lim (\frac{3}{x ^{2} - x})

\int \frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

What is the derivative of ((sqrt(5x^4+x^2))/2) ?
The derivative of ((sqrt(5x^4+x^2))/2) is (10x^2+1)/(2sqrt(5x^2+1))
What is the first derivative of ((sqrt(5x^4+x^2))/2) ?
The first derivative of ((sqrt(5x^4+x^2))/2) is (10x^2+1)/(2sqrt(5x^2+1))