First principles of differentation - Help Please (1 Viewer)

Smile12345 · Jun 11, 2013

Can someone please help me with this question...

For the function f(x)= x^-1 (x to the power of -1)
a) Evaluate f(x)- f(5) / x - 5... When x = 4.99

Thanks heaps....

braintic · Jun 11, 2013

As you have not mentioned limits, it is a simple substitution.

Smile12345 · Jun 11, 2013

braintic said:
As you have not mentioned limits, it is a simple substitution.

Awe yeah... Thanks heaps... Sometimes I think too complex! Do you ever have that trouble?

Braintic... The next part was to evaluate when x = 5.01... I've got that answer... Now I have to use these results to find the derivative of the function at the point where x = 5... How do I do this?

Sy123 · Jun 12, 2013

Smile12345 said:
Awe yeah... Thanks heaps... Sometimes I think too complex! Do you ever have that trouble?

Braintic... The next part was to evaluate when x = 5.01... I've got that answer... Now I have to use these results to find the derivative of the function at the point where x = 5... How do I do this?

First principles:

$$The notation$ \ \ f'(a)$

$$is used to show the value of the derivative at$ \ \ a$

$In this case$ \ \ a=5 \ \ $and by first principles$

$f(a) = \lim_{h\ to 0} \frac{f(a+h)-f(a)}{h}$

The use of x=4.99 and x=5.01 is rather silly, but essentially when you substituted those values into the formula, you are finding the gradient of the chord from x=5 to x=4.9 and so on. And this show give an estimate as to the gradient of the tangent (the derivative)

$$Since$ \ \ f(x) = \frac{1}{x}$

$f'(5) = \lim_{h \to 0} \frac{\frac{1}{5+h} - \frac{1}{5}}{h}$

$f'(5) = \lim_{h \to 0} \frac{\frac{5 - (5+h)}{5(5+h)}}{h} = \lim_{h \to 0} \frac{-h}{5(5+h)} \times \frac{1}{h} = \frac{-1}{25}$

$$So the value of the derivative at$ \ \ x= 5 \ \ $is given by$ \ \ \frac{-1}{25}$

Smile12345 · Jun 12, 2013

Sy123 said:
First principles:

$$The notation$ \ \ f'(a)$

$$is used to show the value of the derivative at$ \ \ a$

$In this case$ \ \ a=5 \ \ $and by first principles$

$f(a) = \lim_{h\ to 0} \frac{f(a+h)-f(a)}{h}$

The use of x=4.99 and x=5.01 is rather silly, but essentially when you substituted those values into the formula, you are finding the gradient of the chord from x=5 to x=4.9 and so on. And this show give an estimate as to the gradient of the tangent (the derivative)

$$Since$ \ \ f(x) = \frac{1}{x}$

$f'(5) = \lim_{h \to 0} \frac{\frac{1}{5+h} - \frac{1}{5}}{h}$

$f'(5) = \lim_{h \to 0} \frac{\frac{5 - (5+h)}{5(5+h)}}{h} = \lim_{h \to 0} \frac{-h}{5(5+h)} \times \frac{1}{h} = \frac{-1}{25}$

$$So the value of the derivative at$ \ \ x= 5 \ \ $is given by$ \ \ \frac{-1}{25}$

Thanks Sy...Yes, sometimes textbook questions can be quite silly, I agree. Thanks again for your explanation...

First principles of differentation - Help Please (1 Viewer)

Smile12345

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braintic

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Smile12345

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Sy123

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Smile12345

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