Derivative of f(x) = 3/x using limit definition of derivative?

2 Answers
Mar 1, 2018

f'(x)=-3/x^2

Explanation:

The definition of a derivative

f'(x)=lim_(h->0)(f(x+h)-f(x))/h

We want to differentiate f(x)=3/x, therefore we seek

f'(x)=lim_(h->0)(3/(x+h)-3/x)/h

Thus

f'(x)=3lim_(h->0)(1/(x+h)-1/x)/h

=3lim_(h->0)(1/(h(x+h))-1/(hx))

=3lim_(h->0)(x-(x+h))/(h(x+h)x)

=3lim_(h->0)(-h)/(h(x+h)x)

=3lim_(h->0)(-1)/((x+h)x)

=3lim_(h->0)(-1)/(x^2+hx)

=-3/x^2

Mar 1, 2018

Given: f(x) = 3/x then f(x+h) = 3/(x+h)

The limit definition of the derivative is:

f'(x) = lim_(h to 0) (f(x+h)-f(x))/h

Substitute f(x) = 3/x and f(x+h) = 3/(x+h):

f'(x) = lim_(h to 0) (3/(x+h)-3/x)/h

Multiply by 1 in the form of (x(x+h))/(x(x+h)):

f'(x) = lim_(h to 0) (x(x+h))/(x(x+h))(3/(x+h)-3/x)/h

Perform the multiplication:

f'(x) = lim_(h to 0) ((3x(x+h))/(x+h)-(3x(x+h))/x)/(h(x(x+h)))

Please observe how this simplifies the numerator by cancellation:

f'(x) = lim_(h to 0) ((3xcancel((x+h)))/cancel((x+h))-(3cancel(x)(x+h))/cancel(x))/(h(x(x+h)))

The limit with the cancelled factors removed:

f'(x) = lim_(h to 0) (3x-3(x+h))/(h(x(x+h)))

Distribute the -3 in the numerator:

f'(x) = lim_(h to 0) (3x-3x-3h)/(h(x(x+h)))

Combine like terms:

f'(x) = lim_(h to 0) (-3h)/(h(x(x+h)))

h/h becomes 1:

f'(x) = lim_(h to 0) (-3)/(x(x+h))

Now, we may let h to 0:

f'(x) = -3/(x(x))

Simplify:

f'(x) = -3/x^2