Question

Using the limit definition, how do you find the derivative of `f (x) = 3x^5 + 4x`?

Answer 1

Apply the limit definition and use some algebra to simplify to find that

`f'(x) = 15x^4+4`

Explanation:

There are two equivalent definitions commonly used for the derivative of a function at a point:

`f'(a) = lim_(h->0)(f(a+h)-f(a))/h`

and

`f'(a) = lim_(x->a)(f(x)-f(a))/(x-a)`

Note that we can show the second as being equivalent to the first by making the substitution `h=x-a` into the first. We will be using the second for this problem:

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`f'(a) = lim_(x->a)(f(x)-f(a))/(x-a)`

`=lim_(x->a)(3x^5+4x-3a^5+4a)/(x-a)`

`=lim_(x->a)(3*(x^5-a^5)/(x-a)+4*(x-a)/(x-a))`

`=lim_(x->a)(3(x^4+x^3a+x^2a^2+xa^3+a^4)+4)`

`=3(a^4+a^4+a^4+a^4+a^4)+4`

`=3(5a^4)+4`

`=15a^4+4`

Thus, we have `f'(x) = 15x^4+4`