What is the derivative of y=-arcsin(5x^2+4) ?

The derivative of y=-arcsin(5x^2+4) is -(10x)/(sqrt(-25x^4-40x^2-15))

What is the first derivative of y=-arcsin(5x^2+4) ?

The first derivative of y=-arcsin(5x^2+4) is -(10x)/(sqrt(-25x^4-40x^2-15))

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derivative of

y = - arcsin (5 x^{2} + 4)

- \frac{10 x}{- 25 x ^{4} - 40 x ^{2} - 15}

Solution steps

\frac{d}{d x} (- arcsin (5 x^{2} + 4))

Take the constant out:

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

= - \frac{d}{d x} (arcsin (5 x^{2} + 4))

Apply the chain rule:

\frac{1}{1 - ( 5 x ^{2} + 4 ) ^{2}} \frac{d}{d x} (5 x^{2} + 4)

= \frac{1}{1 - ( 5 x ^{2} + 4 ) ^{2}} \frac{d}{d x} (5 x^{2} + 4)

\frac{d}{d x} (5 x^{2} + 4) = 10 x

= - \frac{1}{1 - ( 5 x ^{2} + 4 ) ^{2}} \cdot 10 x

Simplify

- \frac{1}{1 - ( 5 x ^{2} + 4 ) ^{2}} \cdot 10 x : - \frac{10 x}{- 25 x ^{4} - 40 x ^{2} - 15}

= - \frac{10 x}{- 25 x ^{4} - 40 x ^{2} - 15}

f (x) = 3^{2 x + 1}

\int_{1}^{4} e^{x} d x

\int x^{2} - 16 d x

f (x) = x^{3} - 2 x + 1, (3, 22)

f (x) = \frac{1 + ln ( x )}{x}, at x = e

What is the derivative of y=-arcsin(5x^2+4) ?
The derivative of y=-arcsin(5x^2+4) is -(10x)/(sqrt(-25x^4-40x^2-15))
What is the first derivative of y=-arcsin(5x^2+4) ?
The first derivative of y=-arcsin(5x^2+4) is -(10x)/(sqrt(-25x^4-40x^2-15))