Easy way to do this derivative??? (Calc 1)?

Question

I'm doing a practice exam for our first Calc 1 test, and question 7 part b) asks: Find f'(x) of y=(x^5 + 4x^3 - x)^6. I know the formula for f'(x) is lim(h->0) [f(x+h)-f(x)]/h, and I can do this problem the traditional way but it would take FOREVER to take all of those x+h's to the different powers, then find the entire f(x+h) to the sixth power and f(x) to the sixth...

Given that this is just a part of one question.. and part a) was easy.. and the test has to be completed within an hour, there must be some exponential rules that allow this to be done much more quickly. Can anyone help?

Thanks!!! Haha, Doug, I know I figured I'd up my odds since most people in the physics forum are as good at calc as those in the math forum... probably better actually given the amount of math that can be done without calc compared to the amount of physics that...can't ;-). Thanks again!

Answer 1

yes, I will help because I hate how they teach calculus.

The general formula goes like this.
D(a*x^n)/dx = n*a*x^(n-1)

Where a is a constant.

But for your case you also have to use the "chain" rule.

I will give an example.

say you have a function like f(x) = (x^2 + x)^2

then you have to take the derivative of the "inside function" which is 2*x +1 in this case (using the first formula I gave) and multiply it by the derivative of the "outside function" to get the total derivative. the derivative of the outside funtion is found by doing it this way: let v = x^2 + x then f(v) = (v)^2

now find f'(v) as 2v using the first formulas.

Now you must multiply the two parts together, hence its called the chain rule.
the answer is:
f'(x) = 2v*(2x+1) = 2(x^2 + x) * (2x + 1)

Good luck!

Answer 2

Well, if you have learned the Chain Rule, then you should use that approach.

The Chain Rule: Let u and v be two differentiable functions. Then (u o v) is differentiable and (u o v)'(x) = v'(x) * u'(v(x)).

So for this problem u(x) = x^6 and v(x) = (x^5 + 4x^3 - x)

Derivatives -- u'(x) = 6x^5 and v'(x) = (5x^4 + 12x^2 -1)

Now applying the Chain Rule -- (u o v)'(x) = v'(x) * u'(v(x)):
= 6 * (5x^4 + 12x^2 -1) * (x^5 + 4x^3 - x)^5

* Remark * -- Let u(x) be a differentiable function. For any real number N, f(x) = [u(x)]^N is differentiable and
f '(x) = N * u'(x) * [u(x)]^N-1.

Answer 3

Chain rule- [6(x^5+4x^3-x)^5] multiplied by [5x^4+12x^2-1]

Answer 4

Chain rule, and power rule.

[6(x^5 + 4x^3 - x^1)](5x^4 + 12x^2 - 1x^0)

Answer 5

Hey Dewd!!!!! I just answered that in the Math Forum ☺

Doug