# Derivative of function with fractional exponent

1. Mar 19, 2006

### tony873004

The book and lecture notes do not give a good example of how to solve this type of problem. After writing out f' I don't know how to simplify. Any hints?
$$\begin{array}{l} f(x) = x - 5(x - 1)^{2/5} \\ \\ f'(x) = \frac{{f(x + h) - f(x)}}{h} = \frac{{(x + h) - 5((x + h) - 1)^{2/5} - \left( {x - 5(x - 1)^{2/5} } \right)}}{h} \\ \end{array}$$

2. Mar 19, 2006

### Hurkyl

Staff Emeritus
First off, f'(x) is not given by:

$$f'(x) = \frac{f(x + h) - f(x)}{h}$$

(that expression doesn't even make sense: where did h come from?)

I know you meant

$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$

but it's important to remember that it's a limit! Sometimes people forget.

As to actually evaluating the derivative, have you learned the algebraic techniques yet? Like the sum rule, the power rule, the product rule, the chain rule, and the like? This would be a very irritating expression to try and evaluate from the limit definition!

3. Mar 19, 2006

### tony873004

lol. I was explaining to a classmate that lim h->0 must go in the problem, and now I forgot myself.

This is in the section before the sum rule, power rule, and chain rule. We are forbidden from using them on problems in this section.

I wouldn't even know how to use the power rule here since the exponent isn't for x, but for (x-1).

4. Mar 19, 2006

### Hurkyl

Staff Emeritus
I can't think of any useful techniques that someone just starting their first calculus class would be expected to know, before encountering the derivative rules!

The only technique I've yet imagined that you'd be able to follow is to rationalize the numerator (much as you would do when trying to differentiate the square root of x), but unless you've been shown the trick for higher roots, I can't imagine you'd be expected to imagine it yourself.

Anyways, if you do it with the algebraic rules, don't forget about the chain rule!

Last edited: Mar 19, 2006
5. Mar 19, 2006

### tony873004

The specific directions were to use technology to graph the deriative, then use the graph to estimate all values of x (if any) where the function is not differentiable, and the tangent line to the graph of the given function is horizontal.

By "Technology", this book means the TI-83+ or TI-84 plus calculator, or Microsoft Excel. I'm not aware if a method exists where I can enter a formula, and it graphs the derivative for me. So I just figured I had to compute it myself. Any thoughts?

6. Mar 19, 2006

### 0rthodontist

7. Mar 20, 2006

### tony873004

Thanks. This makes more sense than doing it by hand.

The 200 page manual scare me away. I can't even find it now. I'll have to search online for it.

I'm pressing MATH 8, and then entering

nDeriv(X-5(X-1)^(2/5))

and I get an ERR:ARGUMENT

To enter X, I'm pressing ALPHA STO>
I've also tried the X,T,theta,n button and I get the same error. Even

nDeriv(3) gives me the error, so I imagine I'm doing something wrong.

8. Mar 20, 2006

### 0rthodontist

1. you want to put it into one of the Y= fields, or else it's not going to graph
2. you also have to specify which variable you would like to differentiate with respect to, and at what point you would like to differentiate it. I recommend something like nDeriv(Z-5(Z-1)^(2/5),Z,X). Bear in mind that I myself use a TI-89 which has different syntax and functions.
3. you should play around with your calculator until you know how to use it, since experimentation is often faster than the manual.

9. Mar 20, 2006

### tony873004

To do this question with "Technology", it would need to graph the derivative function. If I have to tell it what at what point, then it is only giving me the slope at that point, rather than a graph off'(x), right?

Can I enter 3x^2 and have it give me a graph of 6x?, or simply tell me that f'(x)=6x, or do I have to tell it 3x^2 at x=4 which would make it tell me 24.?

10. Mar 21, 2006

### HallsofIvy

Staff Emeritus
I'm using a TI 85 and here's how I would do that: press the graph key, then choose "y= " to edit the function. Instead of directly entering "x^2", I press the "catalog" key and scroll down to "Der1(". Enter "x^2, x" to tell the computer the function to be differentiated and the variable of differentiation, and the proceed as normally to graph that.

11. Mar 21, 2006

### Curious3141

Actually, it's perfectly doable from first principles, as long as you can use the general form of the binomial theorem.

I don't think there's any harm here in showing full working, it should be illustrative.

$$\begin{array}{l} f(x) = x - 5(x - 1)^{2/5} \\ \\ f'(x) =\lim_{h\rarrow 0} \frac{{f(x + h) - f(x)}}{h} = \frac{{(x + h) - 5((x + h) - 1)^{2/5} - \left( {x - 5(x - 1)^{2/5} } \right)}}{h} \\ \end{array}$$
So (I'm going to forego the limit notation here, and just take the limit at the end). $$f(x+h) = (x + h) - 5(x-1 +h)^{\frac{2}{5}}$$
$$f(x+h) = (x + h) - 5(x-1)^{\frac{2}{5}}(1 +\frac{h}{x-1})^\frac{2}{5}$$

Expand that with Binomial Theorem to a first order approximation (the higher order terms will vanish once the limit is taken anyway). I am including the second order term for clarity.

$$f(x+h) = (x + h) - 5(x-1)^{\frac{2}{5}}(1 +(\frac{2}{5})\frac{h}{x-1} - (\frac{3}{25})(\frac{h^2}{(x-1)^2} + ...)$$

Now subtract $f(x)$ from that.
$$f(x+h) - f(x) = (x + h - x) - 5(x-1)^{\frac{2}{5}}(1 +(\frac{2}{5})\frac{h}{x-1} - 1) - (\frac{3}{25})(\frac{h^2}{(x-1)^2} + ...)$$
$$f(x+h) - f(x) = h - 5(x-1)^{\frac{2}{5}}((\frac{2}{5})\frac{h}{x-1} ) - (\frac{3}{25})(\frac{h^2}{(x-1)^2} + ...)$$

Divide that by h :
$$\frac{f(x+h) - f(x)}{h} = 1 - 5(x-1)^{(\frac{2}{5})}(\frac{2}{5}\frac{1}{x-1} ) - (\frac{3}{25})(\frac{h}{(x-1)^2} + ...)$$

Take the limit as h tends to zero, all higher order terms vanish, and you're left with :
$$f'(x) = 1 - 5(x-1)^{\frac{2}{5}}((\frac{2}{5})\frac{1}{x-1}) = 1 - 2(x-1)^\frac{-3}{5}$$

I rushed thru this, there may be a bracketing error or a sign error here and there, but I think you should get the general idea.

Last edited: Mar 21, 2006
12. Mar 21, 2006

### Hurkyl

Staff Emeritus
Which isn't exactly first principles. I don't quite think it's fair to use infinite series before you are even allowed to use the derivative rules! And it isn't something a beginning calc student would be expected to know anyways.

Incidentally, the only way I see to do this using only knowledge available to a beginning calc student is to go from

$$a^{1/5} - b^{1/5}$$

to

$$\frac{(a^{1/5} - b^{1/5})(a^{4/5} + a^{3/5} b^{1/5} + a^{2/5} b^{2/5} + a^{1/5} b^{3/5} + b^{4/5})}{(a^{4/5} + a^{3/5} b^{1/5} + a^{2/5} b^{2/5} + a^{1/5} b^{3/5} + b^{4/5})}$$

which rationalizes the numerator.

13. Mar 21, 2006

### tony873004

It's funny, but the teacher just blindly assigns us problems 1-71 odd, without actually looking at what he is assigning us. When I asked him how to the do the problem, he was also surprised that such a problem would be assigned in this section. I don't have my notes with me, but I'll post his solution later.

HallsOfIvy, I still get a ERR:ARGUMENT. I'm typing:
/Y1=nDeriv(X2,X)

My teacher says the calculator can't graph the derivative, that it can only tell me the value of the deravite for a given value of X.

14. Mar 21, 2006

### 0rthodontist

Tony, I already told you how to fix the err:argument. Y1 = nDeriv(Z^2, Z, X) should graph the derivative of Z^2 with respect to Z, evaluated at X.

15. Mar 21, 2006

### tony873004

Thanks, now I get it.