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

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

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

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

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derivative of

y = 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}

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