Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
![](https://latex.codecogs.com/png.latex?\bg_white $We have$)
![](https://latex.codecogs.com/png.latex?\bg_white $ $\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left( \sin^{-1} \left(\frac{2x-3}{4}\right) \right)&= \frac{1}{\sqrt{16-\left(2x-3\right)^2}}\times 2, \end{align*})
![](https://latex.codecogs.com/png.latex?\bg_white $\noindent using the chain rule and the rule for differentiating $\sin^{-1} \left(\frac{x}{a}\right)$. Now just simplify that expression to get to the answer.$)
Find the derivative of in the simplest form :
sin^-1 1/4 . ( 2x - 3)
Answer says 2/ (7 + 12x - 4x^2)^1/2
Not sure what to do with the 1/4
Do you multiply it with the numbers in the brackets??