It is given that $y = \sin^{-1} px$, where $p$ is a constant. Show that \(\frac{\

Question

It is given that $y = \sin^{-1} px$, where $p$ is a constant. Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = p \sec y$.[2] By further differentiation of the result in part (a) find, in terms of $p$, the Maclaurin series for $y$ up to and including the term in $x^3$. [5] Use your answer to part (b) to find the series expansion of $\frac{1}{\sqrt{1 - 9x^2}}$, up to and including the term in $x^2$. [3]

Accepted Answer

(b) $y=px+\dfrac{p^3x^3}{6}+\cdots$ (c) $\dfrac1{\sqrt{1-9x^2}}=1+\dfrac92x^2+\cdots$

N2025 P1 Q9

N2025 P1 Q9

Video Solution:

Solution 1