Use the limit definition of the derivative to find 𝑓′(π‘₯) if 𝑓(π‘₯) = π‘₯ 10?

Use the limit definition of the derivative to find 𝑓′(π‘₯) if 𝑓(π‘₯) = π‘₯
10. Expand all the way out. I came up with a 10 term answer, not sure if that correct or not.
Thank you!

Ash2021-02-17T19:56:23Z

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𝑓(π‘₯) = π‘₯¹⁰
𝑓'(π‘₯) = lim(h β†’ 0) [𝑓(π‘₯+h) - 𝑓(π‘₯)]/h
𝑓'(π‘₯) = lim(h β†’ 0) [(π‘₯+h)¹⁰ - π‘₯¹⁰]/h
Using Pascal's triangle expand (π‘₯+h)¹⁰
𝑓'(π‘₯) = lim(h β†’ 0) [π‘₯¹⁰ + 10π‘₯⁹h + 45π‘₯⁸hΒ² + 120π‘₯⁷hΒ³ + 210π‘₯⁢h⁴ + 252π‘₯⁡h⁡ + 210π‘₯⁴h⁢ + 120π‘₯Β³h⁷ + 45π‘₯Β²h⁸ + 10π‘₯h⁹ + h¹⁰ - π‘₯¹⁰]/h
𝑓'(π‘₯) = lim(h β†’ 0) [10π‘₯⁹h + 45π‘₯⁸hΒ² + 120π‘₯⁷hΒ³ + 210π‘₯⁢h⁴ + 252π‘₯⁡h⁡ + 210π‘₯⁴h⁢ + 120π‘₯Β³h⁷ + 45π‘₯Β²h⁸ + 10π‘₯h⁹ + h¹⁰]/h
𝑓'(π‘₯) = lim(h β†’ 0) h[10π‘₯⁹ + 45π‘₯⁸h + 120π‘₯⁷hΒ² + 210π‘₯⁢hΒ³ + 252π‘₯⁡h⁴ + 210π‘₯⁴h⁡ + 120π‘₯Β³h⁢ + 45π‘₯Β²h⁷ + 10π‘₯h⁸ + h⁹]/h
𝑓'(π‘₯) = lim(h β†’ 0) [10π‘₯⁹ + 45π‘₯⁸h + 120π‘₯⁷hΒ² + 210π‘₯⁢hΒ³ + 252π‘₯⁡h⁴ + 210π‘₯⁴h⁡ + 120π‘₯Β³h⁢ + 45π‘₯Β²h⁷ + 10π‘₯h⁸ + h⁹]

Now plug in h=0
𝑓'(π‘₯) = 10π‘₯⁹

az_lender2021-02-17T17:43:45Z

No, you don't need 10 terms.

f(x+h) = (x+h)^10.

f(x+h) - f(x) = 10*x^9*h + 45*x^8*h^2 + 120*x^7*h^3 + blah blah blah + h^10.

When you divide this by h, you get

[f(x+h) - f(x)]/h = 10*x^9 + 45*x^8*h + 120*x^7*h^2 + blah blah blah + h^9.

Now, when you take the limit as h-->0, you get only 10x^9.Β  Period!