What is the derivative of 4arcsin((2x-4/(3x+2))) ?

The derivative of 4arcsin((2x-4/(3x+2))) is (64)/((3x+2)sqrt(5x^2+28x-12))

What is the first derivative of 4arcsin((2x-4/(3x+2))) ?

The first derivative of 4arcsin((2x-4/(3x+2))) is (64)/((3x+2)sqrt(5x^2+28x-12))

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\frac{d}{d x} (4 arcsin (\frac{2 x - 4}{3 x + 2}))

\frac{64}{( 3 x + 2 ) 5 x ^{2} + 28 x - 12}

Solution steps

\frac{d}{d x} (4 arcsin (\frac{2 x - 4}{3 x + 2}))

Take the constant out:

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

= 4 \frac{d}{d x} (arcsin (\frac{2 x - 4}{3 x + 2}))

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{2 x - 4}{3 x + 2}) = \frac{16}{( 3 x + 2 ) ^{2}}

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

Simplify

4 \cdot \frac{1}{1 - ( \frac{2 x - 4}{3 x + 2} ) ^{2}} \cdot \frac{16}{( 3 x + 2 ) ^{2}} : \frac{64}{( 3 x + 2 ) 5 x ^{2} + 28 x - 12}

= \frac{64}{( 3 x + 2 ) 5 x ^{2} + 28 x - 12}

x \to 1 lim (\frac{3 - \frac{3}{x}}{x - 1})

e^{20}

y = \frac{12 x}{x ^{2} - 4}, (4, 4)

\frac{x + 2 x t + cos ( t )}{1 + t ^{2}}

5 - x

What is the derivative of 4arcsin((2x-4/(3x+2))) ?
The derivative of 4arcsin((2x-4/(3x+2))) is (64)/((3x+2)sqrt(5x^2+28x-12))
What is the first derivative of 4arcsin((2x-4/(3x+2))) ?
The first derivative of 4arcsin((2x-4/(3x+2))) is (64)/((3x+2)sqrt(5x^2+28x-12))