What is the derivative of sqrt(1+x^2+x^6) ?

The derivative of sqrt(1+x^2+x^6) is (x(3x^4+1))/(sqrt(1+x^2+x^6))

What is the first derivative of sqrt(1+x^2+x^6) ?

The first derivative of sqrt(1+x^2+x^6) is (x(3x^4+1))/(sqrt(1+x^2+x^6))

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\frac{d}{d x} (1 + x^{2} + x^{6})

\frac{x ( 3 x ^{4} + 1 )}{1 + x ^{2} + x ^{6}}

Solution steps

\frac{d}{d x} (1 + x^{2} + x^{6})

Apply the chain rule:

\frac{1}{2 1 + x ^{2} + x ^{6}} \frac{d}{d x} (1 + x^{2} + x^{6})

= \frac{1}{2 1 + x ^{2} + x ^{6}} \frac{d}{d x} (1 + x^{2} + x^{6})

\frac{d}{d x} (1 + x^{2} + x^{6}) = 6 x^{5} + 2 x

= \frac{1}{2 1 + x ^{2} + x ^{6}} (6 x^{5} + 2 x)

Simplify

\frac{1}{2 1 + x ^{2} + x ^{6}} (6 x^{5} + 2 x) : \frac{x ( 3 x ^{4} + 1 )}{1 + x ^{2} + x ^{6}}

= \frac{x ( 3 x ^{4} + 1 )}{1 + x ^{2} + x ^{6}}

y^{^{''}} + y^{^{'}} = 3

f (x) = x^{2} - x

\frac{d}{d s} (4^{- 3} (\frac{2}{s ^{7}} + 3 s)^{4})

\int x^{3} ln ∣ 1 + x ∣ d x

x \to - 1 lim (\frac{x - 3}{x + 1})

What is the derivative of sqrt(1+x^2+x^6) ?
The derivative of sqrt(1+x^2+x^6) is (x(3x^4+1))/(sqrt(1+x^2+x^6))
What is the first derivative of sqrt(1+x^2+x^6) ?
The first derivative of sqrt(1+x^2+x^6) is (x(3x^4+1))/(sqrt(1+x^2+x^6))