Solving a Derivative Puzzle: Find f'(1)

[terryds]

Homework Statement

If $\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1$.
then f'(1) = ...

A. -1
B. -1/3
C. 1/3
D. 1
E. 2

Homework Equations

[/B]
$f'(x) = \lim_{h->0}\frac{f(x+h)-f(x))}{h}$

The Attempt at a Solution

[/B]
I really have no idea about this problem.
Please help me.

 

[PeroK]

Homework Statement

If $\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1$.
then f'(1) = ...

A. -1
B. -1/3
C. 1/3
D. 1
E. 2

Homework Equations

[/B]
$f'(x) = \lim_{h->0}\frac{f(x+h)-f(x))}{h}$

The Attempt at a Solution

[/B]
I really have no idea about this problem.
Please help me.

First, rewrite the definition of the derivative limit. There are always two alternative ways o wrtite a limit. And, I'm not talking about epsilon-delta.

Second, a hint: you need to learn to spot a composite function when you see one!

 

[terryds]

First, rewrite the definition of the derivative limit. There are always two alternative ways o wrtite a limit. And, I'm not talking about epsilon-delta.

Second, a hint: you need to learn to spot a composite function when you see one!

Do you mean the L' Hospital rule?
By using L'Hospital rule, I get
$\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1$
$\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1$
$\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1$
$3a^2 f'(a^3) = -a$
$f'(a^3) = \frac{-a}{3a^2}$
$f'(a^3) = -\frac{1}{3a}$
$f'(1^3) = -\frac{1}{3}$

Is it correct ?

 

[PeroK]

Do you mean the L' Hospital rule?
By using L'Hospital rule, I get
$\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1$
$\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1$
$\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1$
$3a^2 f'(a^3) = -a$
$f'(a^3) = \frac{-a}{3a^2}$
$f'(a^3) = -\frac{1}{3a}$
$f'(1^3) = -\frac{1}{3}$

Is it correct ?

An interesting solution! That's probably valid, but it's not what I meant. Note that:

$f'(a) \equiv \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a} \equiv \lim_{h\rightarrow 0} \frac{f(a + h)-f(a)}{h}$

Those are the two equivalent ways of writing the definition of a derivative.

And, if you let $g(x) = f(x^3)$, then what you were given was effectively $\forall a, \ g'(a) = -1$. The rest follows from the chain rule.

 

[terryds]

An interesting solution! That's probably valid, but it's not what I meant. Note that:

$f'(a) \equiv \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a} \equiv \lim_{h\rightarrow 0} \frac{f(a + h)-f(a)}{h}$

Those are the two equivalent ways of writing the definition of a derivative.

And, if you let $g(x) = f(x^3)$, then what you were given was effectively $\forall a, \ g'(a) = -1$. The rest follows from the chain rule.

g'(a) = -1
3a^2 f'(a^3) = -1
We let a = 1
f'(1)= -1/3
Thanks for your help

 

[mfb]

As the problem is solved, here is a different approach: Define $y=x^3$, $b=a^3$.
Then
$$-1 = \lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} \\
= \lim_{y\rightarrow b} \frac{f(y)-f(b)}{\sqrt[3]{y}-\sqrt[3]{b}} \cdot \frac{\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2}{\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2} \\
= \lim_{y\rightarrow b} \frac{f(y)-f(b)}{y-b} \cdot \left(\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2\right)$$
The fraction goes to f'(b), the second part converges to 3. Plug in b=1, done.

 

[SammyS]

Similarly:

$\displaystyle \lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} \$

$\displaystyle =\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x^3-a^3}\frac{x^3-a^3}{x-a} \$

$\displaystyle =\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x^3-a^3}\cdot \lim_{x\rightarrow a}\frac{(x-a)(x^2+ax+a^2)}{x-a} \$

$\displaystyle =f'(a)\cdot \lim_{x\rightarrow a}(x^2+ax+a^2)\$​