Differentiate the following function
(1/x)-(2/x^2)+(3/x^3)
I got (-9x-3)/x^5
You should be able to do this in your head, if not, something is wrong, all there are the power formula.
f'=-x-2+4x-3-9x-4
check that.
I don't see how you got that ...
let y = (1/x)-(2/x^2)+(3/x^3)
= x^-1 - 2x^-2 + 3x^-3
dy/dx = - x^-2 + 4x^-3 - 9x^-4
= -1/x^2 + 4/x^3 - 9/x^4
= (-x^2 + 4x - 9)/x^4 , if you want a common denominator
To differentiate the given function, let's break it down into individual terms and apply the power rule.
The given function is:
f(x) = (1/x) - (2/x^2) + (3/x^3)
To differentiate each term, we can write it as:
f(x) = (1 * x^(-1)) - (2 * x^(-2)) + (3 * x^(-3))
Now let's differentiate each term using the power rule:
For the term (1 * x^(-1)):
Differentiating x^(-1) using the power rule, we bring down the exponent and subtract 1 from it:
d/dx (1 * x^(-1)) = -1 * x^(-2) = -x^(-2)
For the term (2 * x^(-2)):
Differentiating x^(-2) using the power rule:
d/dx (2 * x^(-2)) = -2 * x^(-3) = -2/x^3
For the term (3 * x^(-3)):
Differentiating x^(-3) using the power rule:
d/dx (3 * x^(-3)) = -3 * x^(-4) = -3/x^4
Now, adding all the differentiated terms together, we get:
f'(x) = (-x^(-2)) - (2/x^3) - (3/x^4)
However, it seems there was an error in your simplification, and the correct simplification would be:
f'(x) = (-x^(-2)) - (2/x^3) - (3/x^4)
Thus, the derivative of the given function is:
f'(x) = (-x^(-2)) - (2/x^3) - (3/x^4)