What is the derivative of $f (x) = \frac{arcsin (- 10 x - 4)}{9}$ $f(x) = arcsin(-10 x - 4)/9$ ?

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Answer 1 · 2018-01-15T15:02:11

$\frac{d y}{d x} = - \frac{10}{9 \sqrt{1 - {(10 x + 4)}^{2}}}$ $(dy)/(dx)=-10/(9sqrt(1-(10x+4)^2)$

As $y = f (x) = \frac{arcsin (- 10 x - 4)}{9}$ $y=f(x)=arcsin(-10x-4)/9$

$arcsin (- 10 x - 4) = 9 y$ $arcsin(-10x-4)=9y$ or $sin (9 y) = - 10 x - 4$ $sin(9y)=-10x-4$

i.e. $sin (9 y) + 10 x + 4 = 0$ $sin(9y)+10x+4=0$

and differentiating

$9 cos (9 y) \frac{d y}{d x} + 10 = 0$ $9cos(9y)(dy)/(dx)+10=0$

or $\frac{d y}{d x} = - \frac{10}{9 cos (9 y)}$ $(dy)/(dx)=-10/(9cos(9y))$

= $- \frac{10}{9 \sqrt{1 - {sin}^{2} (9 y)}}$ $-10/(9sqrt(1-sin^2(9y))$

= $- \frac{10}{9 \sqrt{1 - {(10 x + 4)}^{2}}}$ $-10/(9sqrt(1-(10x+4)^2)$

What is the derivative of f(x) = arcsin(-10 x - 4)/9f(x)=arcsin(−10x−4)9f(x) = arcsin(-10 x - 4)/9?