derivative of 1/x(x+1)

derivative of 1/x(x+1)

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anyone?
mariad
03 August 2012
How much calculus do you know already? Have you had a go at this yourself? Anyway, this is how to do it, which hopefully you can follow. Rewrite the expression as (x^2 + x)^-1. Which is read "x squared plus x, all to the power of -1". Raising something to the power of -1 is the same thing as dividing 1 by that something, so that a^-1 = 1/a. Now we can use the chain rule of differential calculus. Here we have a "function of a function", since we are applying the function "to the power -1" to some other function of x, namely x^2 + x. Use the result that the derviative of x to some power, say n, is: (d/dx)(x^n) = n*x^(n-1) The chain rule now says that if we say that y = (x^2 + x)^-1, and let u = x^2 + x, then: (dy/dx) = (dy/du) * (du/dx) Using my expression for u we find that y = u^-1. So: dy/du = (-1)*u^(-2), and du/dx = 2x + 1 Using the rule in step 3 on x^2, x and x^-1. So dy/dx, ie the derivative we wanted, is (2x + 1) * (-1) * u^(-2). The final step is to replace the u by its value in terms of x. This was x^2 + x, so we have: dy/dx = (-1) * (2x+1) * (x^2+x)^(-2) Or, rearranging slightly and just using - for "(-1)*" : dy/dx = -(2x + 1)/(x(x+1))^2 Hope this is clear and helps.
jim360
03 August 2012
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