What is the derivative of arcsin(4x+1) ?

The derivative of arcsin(4x+1) is 4/(sqrt(-16x^2-8x))

What is the first derivative of arcsin(4x+1) ?

The first derivative of arcsin(4x+1) is 4/(sqrt(-16x^2-8x))

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derivative of

arcsin (4 x + 1)

\frac{4}{- 16 x ^{2} - 8 x}

Solution steps

\frac{d}{d x} (arcsin (4 x + 1))

Apply the chain rule:

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

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

\frac{d}{d x} (4 x + 1) = 4

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

Simplify

\frac{1}{1 - ( 4 x + 1 ) ^{2}} \cdot 4 : \frac{4}{- 16 x ^{2} - 8 x}

= \frac{4}{- 16 x ^{2} - 8 x}

\int (sin (2 x))^{3} (cos (2 x))^{2} d x

\int (1 + x^{2}) e^{- x} d x

n = 0 \sum \infty \frac{n}{3 n ^{3} + 1}

\frac{d y}{d x} = \frac{( x + y )}{3 x + 3 y - 4}

(2 x + 3 y) d x + (3 x + 2 y) d y = 0

What is the derivative of arcsin(4x+1) ?
The derivative of arcsin(4x+1) is 4/(sqrt(-16x^2-8x))
What is the first derivative of arcsin(4x+1) ?
The first derivative of arcsin(4x+1) is 4/(sqrt(-16x^2-8x))