What is the derivative of 1/(sqrt(1+4x^2+9x^4)) ?

The derivative of 1/(sqrt(1+4x^2+9x^4)) is -(2x(9x^2+2))/((1+4x^2+9x^4)^{3/2)}

What is the first derivative of 1/(sqrt(1+4x^2+9x^4)) ?

The first derivative of 1/(sqrt(1+4x^2+9x^4)) is -(2x(9x^2+2))/((1+4x^2+9x^4)^{3/2)}

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d θ} (2 - 2 sin (θ))

\frac{d}{d x} (cos (3 5 x^{2}))

\int \frac{e ^{2 t}}{e ^{2 t} - 1} d t

\frac{d}{d x} (ln (\frac{1 + tan ( x )}{1 + sin ( x )}))

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{19}{2}}} d t

What is the derivative of 1/(sqrt(1+4x^2+9x^4)) ?
The derivative of 1/(sqrt(1+4x^2+9x^4)) is -(2x(9x^2+2))/((1+4x^2+9x^4)^{3/2)}
What is the first derivative of 1/(sqrt(1+4x^2+9x^4)) ?
The first derivative of 1/(sqrt(1+4x^2+9x^4)) is -(2x(9x^2+2))/((1+4x^2+9x^4)^{3/2)}