What is the derivative of #(x^2 + 1/x)^5#?

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If # y=f(x) # then # f'(x)=dy/dx=dy/(du)(du)/dx #

I was taught to remember that the differential can be treated like a fraction and that the "#dx#'s" of a common variable will "cancel" (It is important to realise that #dy/dx# isn't a fraction but an operator that acts on a function, there is no such thing as "#dx#" or "#dy#" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

# dy/dx = dy/(dv)(dv)/(du)(du)/dx # etc, or # (dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx) #

So with # y = (x^2+1/x)^5 #, Then:

# { ("Let "u=x^2+1/x=x^2+x^-1, => , (du)/dx=2x-x^-2=2x-1/x^2), ("Then "y=u^5, =>, dy/(du)=5u^4 ) :}#

Using # dy/dx=(dy/(du))((du)/dx) # we get:

# dy/dx = (5u^4)(2x-1/x^2) #
# :. dy/dx = 5(2x-1/x^2) (x^2+1/x)^4 #

Hello,

So, this isn't as complicated as it looks. Let's not get daunted by the fraction so much, but focus generally on the main concept. First, we must recognize that this is an application of the Chain Rule for derivatives. Consider the chain rule as the derivative of a composite function. The formula for the chain rule can be stated as follows:

Let f(x) and g(x) be two functions. Then, the composite function of #f# with respect to #g# is #h(x) = f(g(x))#. Therefore, the derivative is #h'(x) = f'(g(x)) \cdot g'(x)#. Looks complicated, but let's try to work through the problem.

First, let's set #g(x) = x^2 + \frac{1}{x}#. Therefore, #h(x) = (g(x))^5#. Note that this is the exact same function you have in your problem; I am just rewriting it. Now, #h'(x) = 5(g(x))^4 \cdot g'(x)#. All I did was apply the power rule and multiply by the derivative of the "inside" function. Now, let's find the derivative of g(x)!

Well, the derivative of #x^2# is just #2x# (power rule) and the derivative of #\frac{1}{x}# is #-\frac{1}{x^2}# (power rule, after we set #\frac{1}{x} = x^{-1}#). Therefore, #g'(x) = 2x - \frac{1}{x^2}#.

Now, remember that #g(x) = x^2 + \frac{1}{x}# and we found g'(x). Let's plug all of this back into h'(x)! So we have ...

#h'(x) = 5(x^2 + \frac{1}{x})^4 \cdot (2x - \frac{1}{x^2})#.

Distributing 5 to g'(x), we obtain

#h'(x) = (x^2 + \frac{1}{x})^4 \cdot (10x - \frac{5}{x^2})#.

Feel free to ping me if you have any inquiries!