What is the derivative of sqrt(x^5+4x^2-1) ?

The derivative of sqrt(x^5+4x^2-1) is (5x^4+8x)/(2sqrt(x^5+4x^2-1))

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

The first derivative of sqrt(x^5+4x^2-1) is (5x^4+8x)/(2sqrt(x^5+4x^2-1))

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

\frac{5 x ^{4} + 8 x}{2 x ^{5} + 4 x ^{2} - 1}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (x^{5} + 4 x^{2} - 1) = 5 x^{4} + 8 x

= \frac{1}{2 x ^{5} + 4 x ^{2} - 1} (5 x^{4} + 8 x)

Simplify

\frac{1}{2 x ^{5} + 4 x ^{2} - 1} (5 x^{4} + 8 x) : \frac{5 x ^{4} + 8 x}{2 x ^{5} + 4 x ^{2} - 1}

= \frac{5 x ^{4} + 8 x}{2 x ^{5} + 4 x ^{2} - 1}

k = 2 \sum \infty \frac{1}{k ( ln ( k ) )}

- 6 x

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

\frac{d}{d x} (x csc (x) - cot (x))

n = 0 \sum \infty \frac{6}{2 ^{n}}

What is the derivative of sqrt(x^5+4x^2-1) ?
The derivative of sqrt(x^5+4x^2-1) is (5x^4+8x)/(2sqrt(x^5+4x^2-1))
What is the first derivative of sqrt(x^5+4x^2-1) ?
The first derivative of sqrt(x^5+4x^2-1) is (5x^4+8x)/(2sqrt(x^5+4x^2-1))