What is the derivative of sqrt(3+4x^2+9x^4) ?

The derivative of sqrt(3+4x^2+9x^4) is (2x(9x^2+2))/(sqrt(3+4x^2+9x^4))

What is the first derivative of sqrt(3+4x^2+9x^4) ?

The first derivative of sqrt(3+4x^2+9x^4) is (2x(9x^2+2))/(sqrt(3+4x^2+9x^4))

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\frac{d}{d x} (3 + 4 x^{2} + 9 x^{4})

\frac{2 x ( 9 x ^{2} + 2 )}{3 + 4 x ^{2} + 9 x ^{4}}

Solution steps

\frac{d}{d x} (3 + 4 x^{2} + 9 x^{4})

Apply the chain rule:

\frac{1}{2 3 + 4 x ^{2} + 9 x ^{4}} \frac{d}{d x} (3 + 4 x^{2} + 9 x^{4})

= \frac{1}{2 3 + 4 x ^{2} + 9 x ^{4}} \frac{d}{d x} (3 + 4 x^{2} + 9 x^{4})

\frac{d}{d x} (3 + 4 x^{2} + 9 x^{4}) = 36 x^{3} + 8 x

= \frac{1}{2 3 + 4 x ^{2} + 9 x ^{4}} (36 x^{3} + 8 x)

Simplify

\frac{1}{2 3 + 4 x ^{2} + 9 x ^{4}} (36 x^{3} + 8 x) : \frac{2 x ( 9 x ^{2} + 2 )}{3 + 4 x ^{2} + 9 x ^{4}}

= \frac{2 x ( 9 x ^{2} + 2 )}{3 + 4 x ^{2} + 9 x ^{4}}

\int_{0}^{1} 2 π x (2 e^{- x^{2}}) d x

\int_{0}^{π} t cos (2 t) d t

\int \frac{( x - 2 ) ( x + 3 )}{x x} d x

\int ln (x^{7}) d x

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

What is the derivative of sqrt(3+4x^2+9x^4) ?
The derivative of sqrt(3+4x^2+9x^4) is (2x(9x^2+2))/(sqrt(3+4x^2+9x^4))
What is the first derivative of sqrt(3+4x^2+9x^4) ?
The first derivative of sqrt(3+4x^2+9x^4) is (2x(9x^2+2))/(sqrt(3+4x^2+9x^4))