Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. **There will be no changes to other Yahoo properties or services, or your Yahoo account.** You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

# Find the derivative of the function using the definition of derivative. f(x) = 2x^4?

### 8 Answers

- husoskiLv 72 months agoFavorite Answer
f(x + h) = 2(x + h)^4 = 2(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4)

= 2x^4 + 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

f(x + h) - f(x) = f(x + h) = 2x^4 = 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

[f(x + h) - f(x)] / h = 8x^3 + 12x^2h + 8xh^2 + 2h^3

f'(x) = lim h-->0 [f(x + h) - f(x)] / h = 8x^3

All the other terms have a factor of h and vanish as h-->0.

- 2 months ago
the answer is 8x^3 because

2x^4

bring 4 to the front so:

4(2)x^

is 8x^ now:

now the exponent you subtract one from all exponents:

4-1=3 so

8x^3

- PuzzlingLv 72 months ago
The definition of a derivative is:

. . . . . f(x + h) - f(x)

lim . . -----------------

h->0 . . . . . h

You have:

f(x) = 2x^4

You can also figure out that:

f(x + h) = 2(x + h)^4

The hard part is to now expand that binomial raised to the 4th power. I simply used the expansion for (a + b)^n = C(n,0) a^n b^0 + C(n,1) a^(n-1) b^1 + ... + C(n,n) a^0 b^n.

Or you can expand it all out by distributing.

f(x + h) = 2(x + h)(x + h)(x + h)(x + h)

f(x + h) = 2(x² + 2hx + h²)(x² + 2hx + h²)

f(x + h) = 2[x²(x² + 2hx + h²) + 2hx(x² + 2hx + h²) + h²(x² + 2hx + h²)]

etc.

However you do it, you'll get to:

f(x + h) = 2[x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4]

f(x + h) = 2x^4 + 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

When you subtract f(x), the first term disappears:

f(x + h) - f(x) = 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

Next divide it by h:

[f(x + h) - f(x)]/h = 8x^3 + 12x^2h + 8xh^2 + 2h^3

When you take the limit as h approaches 0, the last 3 terms disappear.

Answer:

8x^3

- PinkgreenLv 72 months ago
f '(x)=

limit {2[(x+h)^4-x^4]/h}, where h is a small increment in x.

h->0

=

limit {2[(x+h)^2+x^2][(x+h)^2-x^2]/h}

h->0

=

limit {2[(x+h)^2+x^2][2hx+h^2]/h}

h->0

=

2[x^2+x^2][2x]

=

8x3

- ?Lv 72 months ago
Recall that the definition is [f(x + h) - f(x)] / h

as h approaches zero; it's a limit.

So, what if f(x) ?

What is f(x+h).

Subtracting these & simplifying gives your numerator.

Divide it all now by h & simplify.

What do you get?

Then evaluate this as h gets very small.

That's your answer! Do do these steps!

Don't forget to vote me best answer for being the first to correctly walk you through without spoiling the answer. That way it gives you a chance to work at it and to get good at it!