if y= sin inverse [ 5x +12 root 1-x2 by 13 find dy by dx - Maths - Continuity and Differentiability

Question

if y= sin inverse [ 5x +12 root 1-x2 by 13 find dy by dx

Shridhar Kaushik · Accepted Answer

We have,

$y = \sin^{- 1} [\frac{5 x + 12 \sqrt{1 - x^{2}}}{13}] \Rightarrow y = \sin^{- 1} [\frac{5}{13} x + \frac{12}{13} \sqrt{1 - x^{2}}]$

Put x = cos θ

$\Rightarrow \sqrt{1 - x^{2}} = \sqrt{1 - \cos^{2} θ} = \sin θ ∴ y = \sin^{- 1} (\frac{5}{13} \cos θ + \frac{12}{13} \sin θ) Now, (\frac{5}{13})^{2} + (\frac{12}{13})^{2} = \frac{144 + 25}{169} = 1$

Let cos α = $\frac{12}{13}$ such that α lies in first quadrant.

$∴ \sin^{2} α = 1 - (\frac{12}{13})^{2} = 1 - \frac{144}{169} = \frac{25}{169} \Rightarrow \sin α = \pm \frac{5}{13} = \frac{5}{13} [In first quadrant, sin α is positive] ∴ y = \sin^{- 1} (\sin α . \cos θ + \cos α \sin θ) = \sin^{- 1} [\sin (α + θ)] = α + θ = \cos^{- 1} (\frac{12}{13}) + \cos^{- 1} x ∴ \frac{d y}{d x} = 0 + (\frac{- 1}{\sqrt{1 - x^{2}}}) = \frac{- 1}{\sqrt{1 - x^{2}}}$

147