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#

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:
#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#
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Answer 2

To find the derivative of ( f(x) = 3x^5 + 4x ) using the limit definition of a derivative, you use the formula:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

Substitute the function ( f(x) = 3x^5 + 4x ) into the formula and simplify the expression.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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