What is the derivative of w/(sqrt(1-4x^2)) ?

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

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

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

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

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

Solution steps

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

Treat

w

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= w (- \frac{1}{2 ( 1 - 4 x ^{2} ) ^{\frac{3}{2}}} (- 8 x))

Simplify

w (- \frac{1}{2 ( 1 - 4 x ^{2} ) ^{\frac{3}{2}}} (- 8 x)) : \frac{4 w x}{( 1 - 4 x ^{2} ) ^{\frac{3}{2}}}

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

2 t^{- \frac{1}{4}} + 2 ln (t) + ln (5)

y^{^{''}} - y^{^{'}} + 0.25 y = 0

\int e^{4 x} sin (b x) d x

f (x) = x + 9

\int 5 x^{\frac{3}{2}} d x

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