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lara31 [8.8K]
3 years ago
6

Let f(X) = (x^7(x-6)^8)/(x^2+5)^9

Mathematics
1 answer:
Semenov [28]3 years ago
5 0
First a review of log properties:
log (ab) = log(a) + log(b)
log (a/b) = log(a) - log(b)
log (a^n) = n*log(a)

In this way by taking the log of entire function we can break it up into more simple terms before trying to take the derivative.

But we must take log of BOTH sides.
y = \frac{x^7 (x-6)^8}{(x^2+5)^9}   \\ ln (y) = ln (\frac{x^7 (x-6)^8}{(x^2+5)^9})
Expand Right side
ln(y) = ln (x^7) + ln (x-6)^8 - ln (x^2+5)^9  \\ ln (y) = 7 ln (x) +8 ln (x-6) - 9 ln(x^2 +5)
Now lets review derivative of natural log:
\frac{d}{dx} ln( u(x)) = \frac{1}{u(x)} \frac{du}{dx}
Take derivative of both sides:
\frac{1}{y} \frac{dy}{dx} = 7(\frac{1}{x})(1) + 8(\frac{1}{x-6})(1) - 9(\frac{1}{x^2+5})(2x)  \\ \frac{1}{y} \frac{dy}{dx} = \frac{7}{x} + \frac{8}{x-6} - \frac{18x}{x^2+5}
Solve for dy/dx by multiplying by "y". (Note y is original function)
\frac{dy}{dx} = y(\frac{7}{x} + \frac{8}{x-6} - \frac{18x}{x^2+5}) \\ \frac{dy}{dx} = (\frac{x^7 (x-6)^8}{(x^2+5)^9})(\frac{7}{x} + \frac{8}{x-6} - \frac{18x}{x^2+5})
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Answer:

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The confidence interval for the population mean x can be calculated as:

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A

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