What is the derivative of y= 7/(3-4x^4) ?

The derivative of y= 7/(3-4x^4) is (112x^3)/((3-4x^4)^2)

What is the first derivative of y= 7/(3-4x^4) ?

The first derivative of y= 7/(3-4x^4) is (112x^3)/((3-4x^4)^2)

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derivative of

y = \frac{7}{3 - 4 x ^{4}}

\frac{112 x ^{3}}{( 3 - 4 x ^{4} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{7}{3 - 4 x ^{4}})

Take the constant out:

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

= 7 \frac{d}{d x} (\frac{1}{3 - 4 x ^{4}})

Apply exponent rule:

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

= 7 \frac{d}{d x} ((3 - 4 x^{4})^{- 1})

Apply the chain rule:

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

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

\frac{d}{d x} (3 - 4 x^{4}) = - 16 x^{3}

= 7 (- \frac{1}{( 3 - 4 x ^{4} ) ^{2}} (- 16 x^{3}))

Simplify

7 (- \frac{1}{( 3 - 4 x ^{4} ) ^{2}} (- 16 x^{3})) : \frac{112 x ^{3}}{( 3 - 4 x ^{4} ) ^{2}}

= \frac{112 x ^{3}}{( 3 - 4 x ^{4} ) ^{2}}

(x + 1) y^{^{'}} + (2 x - 1) y = e^{- 2 x}

\frac{d}{d x} (- 6 sin (6 x))

\frac{\partial}{\partial x} (ln (cos (x^{2} y^{2} + z)))

arcsin (\frac{x}{7}), at x = 4

\frac{d}{d x} (ln (8 - 9 x))

What is the derivative of y= 7/(3-4x^4) ?
The derivative of y= 7/(3-4x^4) is (112x^3)/((3-4x^4)^2)
What is the first derivative of y= 7/(3-4x^4) ?
The first derivative of y= 7/(3-4x^4) is (112x^3)/((3-4x^4)^2)