**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 10

*x*e^{2x}is twice of 5*x*e^{2x}, and this allows the simplification in [1]. The second differentiation yields 10e^{2x}which, we notice, is twice of 5e^{2x}. We can get rid of that term. Multiplying equation [1] by 2 gives 10e^{2x}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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