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secondary 3 | A Maths
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Daddy
Daddy

secondary 3 chevron_right A Maths chevron_right Singapore

Pls help me continue from here

Date Posted: 2 years ago
Views: 210
Eric Nicholas K
Eric Nicholas K
2 years ago
Your expression in the numerator looks wrong…
Eric Nicholas K
Eric Nicholas K
2 years ago
Shouldn’t it be (x - 1) times “0” minus 1?

Differentiating 1 gets us 0.
Eric Nicholas K
Eric Nicholas K
2 years ago
Also, your presentation is incorrect. On the second line of your working you should write dy/dx = … because you have already proceeded to perform the differentiation in the second step.

Anyway, you will get dy/dx = -1 divided by (x - 1)^2.

Note that the curve is not defined for x = 1 since y = 1/(x - 1) = 1/0 has no value.

For all other values of x, (x - 1)^2 will always be a positive value. This is an important property - the output of a squared function is always non-negative.

But the numerator in dy/dx contains a negative sign, so the entire expression -1 divided by (x - 1)^2 is going to be negative in value (negative/positive = negative).

As such, for x < 1, dy/dx is negative…
and for x > 1, dy/dx is also negative.

A negative value for dy/dx indicates that the curve is decreasing in a given interval.

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Arnold K H Tan
Arnold K H Tan's answer
2178 answers (A Helpful Person)
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For this case... differentiate using either the product rule [use y = (x-1)^-1 ] or quotient rule.

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