How do I differentiate y=x^(-2/5) using the power rule?

[nick850]

Homework Statement

Differentiate the function:
y = x ^(-2/5)

Homework Equations

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

The Attempt at a Solution

the -2/5 exponent is giving me a hard time so all i have so far is ((x+h)^(-2/5)-x^(-2/5))/h

 

[stevenb]

Homework Statement

Differentiate the function:
y = x ^(-2/5)

Homework Equations

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

The Attempt at a Solution

the -2/5 exponent is giving me a hard time so all i have so far is ((x+h)^(-2/5)-x^(-2/5))/h

In one approach, the hint would be to think about the formula for binomial series expansion of (x+h)^(-2/5). This is an infinite series, but in the limit as h goes to zero, you won't need all terms.

If you don't know binomial series yet, now is a good time to learn. It gets used over and over and over in different contexts.

 

[nick850]

no i don't know what that is. from the looks from it on wikipedia it looks pretty complicated. I'm sure it's a better way of doing it but i guess my question was more on how to algebraically manipulate the equation so that the limit isn't indeterminate (0/0)

 

[logarithmic]

$$(x+h)^{-2/5} =\,^5\sqrt{(x+h)^2}$$.

To expand the expression in the root, you don't need to fully use the binomial theorem. Then just bring the h inside the root too, and take the limit.

 

[stevenb]

no i don't know what that is. from the looks from it on wikipedia it looks pretty complicated. I'm sure it's a better way of doing it but i guess my question was more on how to algebraically manipulate the equation so that the limit isn't indeterminate (0/0)

As I said, if you don't know the Binomial series expansion, then now is the time to learn it. It's just too important to not know it. It's really not that complicated, but it might look bad symbolically, at first.

I don't know if it really is the best method to solve this problem but it does work very quickly in this case. You end up keeping two terms of an infinite series (other terms go to zero with high orders of h). This limiting form basically allows you to do the manipulation you are looking for, and you can easily remove the indeterminate form.

Usually we use L'Hospital's rule in indeterminate cases, but since that involves taking derivatives, you end up with a circular problem where you need to take the derivative of exactly the same form.

Anyway, rather than leave you hanging, I'll reduce the Binomial Series to the end form you would have (with apologies for writing some sloppy math in the interest of motivation). Maybe this will inspire you to bite the bullet and learn it now.

$$(x+h)^n\approx x^n+nx^{n-1}h$$ when h is small.

If you use this on (x+h)^(-2/5) in the expression ((x+h)^(-2/5)-x^(-2/5))/h you will end up with very simple algebra.

When you get the final answer, you should compare it to the well known derivative rule for powers $${{dx^n}\over{dx}}=nx^{n-1}$$. If fact, you may just want to derive this rule directly and then use it to solve your special case.

 

[Mark44]

Homework Statement

Differentiate the function:
y = x ^(-2/5)

Homework Equations

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

The Attempt at a Solution

the -2/5 exponent is giving me a hard time so all i have so far is ((x+h)^(-2/5)-x^(-2/5))/h

Are you required to use the limit definition of the derivative in this problem? If not, there's a differentiation rule that would make this much easier.

If you have to use the limit definition, you'll need to continue the work started above. It's helpful to get rid of the negative exponents, so your difference quotient can be written as
$$\frac{x^{2/5} - (x + h)^{2/5}}{hx^{2/5}(x + h)^{2/5}}$$

which is equal to
$$\frac{(x^2)^{1/5} - ((x + h)^2)^{1/5}}{hx^{2/5}(x + h)^{2/5}}$$

Before continuing, please verify that you do need to use the limit definition of the derivative.

 

[nick850]

i learned a new trick today which makes it way easier! once again though thank you for your help

 

[Mark44]

The Power Rule? That's why I asked whether you were required to use the limit definition.