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?

Nooo we didn't learn the chain rule, that makes so much sense! Thanks!

The test isn't for another 2 weeks so she's probably going to cover that between now and then...

Answer 1

Sure. If f(x) = g(h(x)) then f'(x) = g'(h(x))*h'(x) so this problem becomes
d/dx (x^5 + 4x^3 - x)^6 = 5(x^5 + 4x^3 - x)^5 * d/dx (x^5 + 4x^3 - x) =
5(x^5 + 4x^3 - x)^5 * (5x^4 + 12x^2 -1)
Haven't you guys done the Chain Rule yet?

Doug

Answer 2

properly, in case you have discovered the Chain Rule, then you somewhat would desire to apply that approach. The Chain Rule: enable u and v be 2 differentiable purposes. Then (u o v) is differentiable and (u o v)'(x) = v'(x) * u'(v(x)). So for this difficulty u(x) = x^6 and v(x) = (x^5 + 4x^3 - x) Derivatives -- u'(x) = 6x^5 and v'(x) = (5x^4 + 12x^2 -a million) Now employing the Chain Rule -- (u o v)'(x) = v'(x) * u'(v(x)): = 6 * (5x^4 + 12x^2 -a million) * (x^5 + 4x^3 - x)^5 * remark * -- enable u(x) be a differentiable function. For any actual style N, f(x) = [u(x)]^N is differentiable and f '(x) = N * u'(x) * [u(x)]^N-a million.

Answer 3

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