How do you find the derivative of $y = arcsin (5 x)$ $y = arcsin(5x)$ ?

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Answer 1 · 2016-11-15T17:22:12

$\frac{d y}{d x} = \frac{5}{\sqrt{1 - 25 x^{2}}}$ $dy/dx = 5/sqrt(1-25x^2)$

$y = arcsin (5 x)$ $y = arcsin(5x)$

$sin y = 5 x$ $siny = 5x$ ..... [1]

We can now differentiate implicitly to get:

$cos (y) \frac{d y}{d x} = 5$ $cos(y)dy/dx = 5$ ..... [2]

Using the fundamental trig identity ${sin}^{2} A + {cos}^{2} A \equiv 1$ $sin^2A+cos^2A-=1$ we can write:

${sin}^{2} (y + {cos}^{2} (y) = 1$ $sin^2(y+cos^2(y)=1$
$:. (5x)^2+cos^2(y)=1$ (from [1])
$:. cos^2(y)=1-25x^2$
$:. cos(y)=sqrt(1-25x^2)$

Substituting into [2] we get:

$sqrt(1-25x^2)dy/dx = 5$

$:. dy/dx = 5/sqrt(1-25x^2)$

How do you find the derivative of y = arcsin(5x)y=arcsin(5x)y = arcsin(5x)?