Question

Find dy/dx when y = 1/x^2-9x? And what is x coordinate at stationary point?

Could you please go through step by step so I understand. Thanks :-)

Answers

f(x) = 1/(x^2 - 9x). Let u = x^2 - 9x.

Our function is now 1/u.

h(x) = 1/x

f(x) = h(u(x)) = 1/u

Let us apply the chain rule:

f'(x) = h'(u(x)) * u'(x)

f'(x) = -1/u^2 * u'

f'(x) = -1/(x^4 - 18x^3 + 81x^2) * (2x - 9)

f'(x) = -(2x - 9)/(x^4 - 18x^3 + 81x^2)

And now, for the stationary point:

f'(x) = 0

0 = -(2x - 9)/(x^3 - 18x^3 + 81x^2)

Multiply both sides by the denominator:

0 = -2x + 9

Solve for x using simple algebra:

2x = 9

x = 9/2

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