Standard +0.3 This is a straightforward application of the chain rule to differentiate an inverse trigonometric function, requiring knowledge of d/dx[arcsin(u)] = 1/√(1-u²) and substitution of a specific x-value. While it requires Further Maths content (inverse trig derivatives), the execution is mechanical with no problem-solving insight needed, making it slightly easier than average overall.
2 In this question you must show detailed reasoning.
Find the gradient of the curve \(y = 6 \arcsin ( 2 x )\) at the point with \(x\)-coordinate \(\frac { 1 } { 4 }\). Express the result in the form \(\mathrm { m } \sqrt { \mathrm { n } }\), where \(m\) and \(n\) are integers.
Question 2:
2 | DR
d y 1
= 6 2
d x 1 − 4 x 2
d y 1
w h e n x = 14 , = 6 2 = 8 3
d x 34 | B1
M1
A1
A1
[4] | 1.1
1.1
1.1
1.1 | 𝑘
√1−4𝑥2
Chain rule (x 2)
𝑑𝑦
Fully correct
𝑑𝑥
Alternative solution | 𝑑𝑥 𝜋
For = 𝑘cos
𝑑𝑦 6
1
Using chain rule (× )
6
d y 1
Or = 6 2
d x 1 − 4 x 2
DR
𝑦
sin = 2𝑥
6
M1
𝑑𝑥 1 𝑦
= cos
𝑑𝑦 12 6 | M1
𝑑𝑦 12 1
= and 𝑥 = ,𝑦 = 𝜋
𝑦
𝑑𝑥 cos 4
6
A1
𝑑𝑦
→ = 8√3
𝑑𝑥
A1
𝑑𝑥 𝜋
For = 𝑘cos
𝑑𝑦 6
1
Using chain rule (× )
6
d y 1
Or = 6 2
d x 1 − 4 x 2
2 In this question you must show detailed reasoning.\\
Find the gradient of the curve $y = 6 \arcsin ( 2 x )$ at the point with $x$-coordinate $\frac { 1 } { 4 }$. Express the result in the form $\mathrm { m } \sqrt { \mathrm { n } }$, where $m$ and $n$ are integers.
\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q2 [4]}}