OCR Further Pure Core 1 2024 June — Question 1 3 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2024
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeDifferentiate inverse trigonometric functions
DifficultyModerate -0.5 This is a straightforward application of the chain rule with the standard derivative of arcsin. While it's a Further Maths topic (making it slightly harder than typical A-level), it requires only direct recall of the formula d/dx[sin⁻¹(u)] = 1/√(1-u²) and basic chain rule application, making it easier than average overall.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08g Derivatives: inverse trig and hyperbolic functions

1 Given that \(y = \sin ^ { - 1 } \left( x ^ { 2 } \right)\), find \(\frac { d y } { d x }\).

Question 1:
AnswerMarks
11
(1−(x2)2
 2 x
 d y  2 x
=
AnswerMarks
d x 1 − x 4B1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.11
For seen.
(1−(x2)2
1 1
For 2 x  f ( x ) where f ( x ) = or ONLY.
1 − ( x 2 ) 2 1−x2
Allow any equivalent correct form e.g. 2 x (1 − x 4 ) − 0 .5 .
Must be in terms of x.
Condone ( x 2 ) 2 for x4.
ISW once correct answer seen.
Alternative method
dy d x
cosy =2x or c o s y = 2 x
AnswerMarks Guidance
dx d yB1 B1
d y
1 − ( x 2 ) 2 = 2 x
AnswerMarks Guidance
d xM1 Replacing c o s y with  1(x2)2 in their derivative of the form
d y
 c o s y = 2 x (or equivalent if differentiating with respect to y).
d x
 d y  2 x
=
AnswerMarks Guidance
d x 1 − x 4A1 Allow any equivalent correct form e.g. 2 x (1 − x 4 ) − 0 .5 .
Must be in terms of x.
Condone ( x 2 ) 2 for x 4 .
ISW once correct answer seen.
[3]
Allow any equivalent correct form e.g. 2 x (1 − x 4 ) − 0 .5 .
Must be in terms of x.
Condone ( x 2 ) 2 for x 4 .
ISW once correct answer seen.
Question 1:
1 | 1
(1−(x2)2
 2 x
 d y  2 x
=
d x 1 − x 4 | B1
M1
A1
[3] | 1.1
1.1
1.1 | 1
For seen.
(1−(x2)2
1 1
For 2 x  f ( x ) where f ( x ) = or ONLY.
1 − ( x 2 ) 2 1−x2
Allow any equivalent correct form e.g. 2 x (1 − x 4 ) − 0 .5 .
Must be in terms of x.
Condone ( x 2 ) 2 for x4.
ISW once correct answer seen.
Alternative method
dy d x
cosy =2x or c o s y = 2 x
dx d y | B1 | B1 | For correctly differentiating implicitly with respect to either x or y. | For correctly differentiating implicitly with respect to either x or y.
d y
1 − ( x 2 ) 2 = 2 x
d x | M1 | Replacing c o s y with  1(x2)2 in their derivative of the form
d y
 c o s y = 2 x (or equivalent if differentiating with respect to y).
d x
 d y  2 x
=
d x 1 − x 4 | A1 | Allow any equivalent correct form e.g. 2 x (1 − x 4 ) − 0 .5 .
Must be in terms of x.
Condone ( x 2 ) 2 for x 4 .
ISW once correct answer seen.
[3]
Allow any equivalent correct form e.g. 2 x (1 − x 4 ) − 0 .5 .
Must be in terms of x.
Condone ( x 2 ) 2 for x 4 .
ISW once correct answer seen.
1 Given that $y = \sin ^ { - 1 } \left( x ^ { 2 } \right)$, find $\frac { d y } { d x }$.

\hfill \mbox{\textit{OCR Further Pure Core 1 2024 Q1 [3]}}