What is the derivative of ${sin}^{- 1} \cdot (5 x)$ $sin^-1 * (5x)$ ?

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Answer 1 · 2017-05-03T13:10:25

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

$y = {sin}^{- 1} (5 x)$ $y=sin^(-1)(5x)$

$\Rightarrow 5 x = sin y$ $=>5x=siny$

differentiate $w r t y$ $" " wrt" " y$

$5 \frac{d x}{d y} = cos y$ $5(dx)/(dy)=cosy$

$\frac{d x}{d y} = \frac{cos y}{5}$ $(dx)/(dy)=cosy/5$

$\Rightarrow \frac{d y}{d x} = \frac{5}{cos y}$ $=>(dy)/(dx)=5/cosy$

we need to rewrite this in terms of the original variable $x$ $x$

now:

${sin}^{2} y + {cos}^{2} y = 1$ $sin^2y+cos^2y=1$

$\Rightarrow cos y = \sqrt{1 - {sin}^{2} y}$ $=>cosy=sqrt(1-sin^2y)$

$\Rightarrow cos y = \sqrt{1 - {(5 x)}^{2}}$ $=>cosy=sqrt(1-(5x)^2)$

$:.cosy=sqrt(1-25x^2)$

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

What is the derivative of sin^-1 * (5x)sin−1⋅(5x)sin^-1 * (5x)?