Calculus problem- basic differentiation rules?

Question

Use first principles to verify the power rule for the case n= -1, that is, if f(x)=1/x, then f'(x)= - 1/x^2 .

does everyone know how to do this?

i mean anyone

Answer 1

Go back to the definition of the derivative:

f'(x) = [h→0]lim (f(x+h) - f(x))/h

In the case where f(x) = 1/x:

[h→0]lim (1/(x+h) - 1/x)/h

Finding a common denominator:

[h→0]lim (x/(x(x+h)) - (x+h)/(x(x+h)))/h

Combining fractions:

[h→0]lim (x-(x+h))/(xh(x+h))

Simplifying:

[h→0]lim -h/(xh(x+h))

Canceling:

[h→0]lim -1/(x(x+h))

Evaluating at h=0:

-1/x²

And we are done.

Answer 2

I'm going to try to err on the side of caution and give you too much information.

The principle beind every derivative is the basic formula for slope, right? (y2-y1)/(x2-x1)=slope. So for any function, the slope for any section of that function can be found using the two endpoints of a segment of the function. The coordinates of these endpoints, simply enough, are (x,f(x)) and (x+h, f(x+h)). You plug these into the slope formula and find that the slope is

(f(x+h)-f(x))/h

and since the derivative is about the slope at a point rather than an average slope, you want the distance between the endpoints here to be zero. That is, you want the limit of the slope as h approaches zero. But if you just plug in zero for h, you have an undefined fraction. So you use your function in the formula and try to simplify, so you can somehow get the h out of the denominator. Like this:

((x+h)^(-1) - x^(-1))/h

Multiply top and bottom by x(x+h)

(x-(x+h))/(hx(x+h))

-h/(hx(x+h))

and cancel the h's

-1/(x(x+h))

And now it's safe to plug in the 0 for h.

-1/(x^2)

Problem solved. Hopefully with clarity and completeness.

Answer 3

the guideline for differentiating a*x^b with admire to x is f(x) = a*x^b -> f'(x) = a*b*x^(b-a million) Differentation works throughout addition like the distributive belongings So on your 0.33 occasion f(x) = 3x^(-2/3)+x^(3/4) f'(x) = 3*(-2/3)*x^(-2/3 - a million) + (3/4)*x^(3/4 - a million) = -2*x^(-5/3) + (3/4)*x^(-a million/4)

Answer 4

Yes.

Use limit definition of the derivative:

f ' (x) = lim {h ---> 0} [f(x + h) - f(x)] / [h]

[f(x + h) - f(x)] / [h] = [ (1 / (x + h)) - (1 / x) ] / h
= [ x - (x + h) / (x)(x + h) ] / h
=[ -h / (x^2 + xh) ] /h
= [ -1 / (x^2 + xh) ]

Now, as h ---> 0, [ -1 / (x^2 + xh) ] = [ -1 / (x^2 + x(0)) ]
= -1 / x^2

Answer 5

d(cx^p)=cpx^p-1)
d(3x^-1)=-3x^-2
d(- 1/x^2)=d(-1x^-2)=2x^-3=2/x^3