Question
Introduction
Although
this looks like a differential equation question, the student is not required
to solve the differential equation. The
requirement is just to derive the equation.
This would be a challenging question for secondary 4 (~ grade 10)
students taking Additional Mathematics or their counterparts in Integrated
Programme schools.
Strategy
One way to
do this is to differentiate the given equation once and again and just verify
the equation by substitution. The
problem is that when we repeatedly apply the Product rule
the terms tend to sprawl. A way to keep things neat is to try to
recognise chunks and also use elimination.
Solution
Remarks
After
differentiating once, we notice that 10xe2x is twice of 5xe2x,
and this allows the simplification in [1]. The second differentiation yields 10e2x which, we notice, is twice of 5e2x. We can get rid of that term. Multiplying equation [1] by 2 gives 10e2x in equation [3], which matches nicely with the same term in [2].
So we can eliminate that term via elimination. After that, we just need to rearrange things
to get the final equation.
H04. Look for pattern(s)
H05. Work backwards
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a
Mathematical Sentence
Suitable Levels
* GCE ‘O’ Level Additional Mathematics, “Integrated Programme
Mathematics”
* GCE ‘A’ Levels H2 Mathematics (revision)
* AP Calculus AB / BC (revision)
* University / College calculus (revision)
* other syllabuses that involve differentiation
* any learner interested in calculus
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