Finding Derivative Using 'Definition of Derivative'

  • #1
CallMeShady
45
1

Homework Statement


mv3l1y.jpg



Homework Equations


As seen above in question.


The Attempt at a Solution


Well, I substituted f(x) into the definition of derivative equation and then multiplied the expression by its conjugate, thus getting the expression in the following form:
f'(a) = lim (a + h)2/3 - a2/3 / h [(a + h)1/3 + a1/3]
...h→0

Now, from this point on, it's just algebra and I am having trouble manipulating this expression so that I can divide the "h" from the denominator. I know I am supposed to use the "hint" given in the question but how would I incorporate that into finding the answer?


Thanks.
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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Homework Statement


mv3l1y.jpg


Homework Equations


As seen above in question.

The Attempt at a Solution


Well, I substituted f(x) into the definition of derivative equation and then multiplied the expression by its conjugate, thus getting the expression in the following form:
f'(a) = lim (a + h)2/3 - a2/3 / h [(a + h)1/3 + a1/3]
...h→0

Now, from this point on, it's just algebra and I am having trouble manipulating this expression so that I can divide the "h" from the denominator. I know I am supposed to use the "hint" given in the question but how would I incorporate that into finding the answer?

Thanks.
Well, multiplying by the conjugate, doesn't work here because getting a difference of squares isn't helpful when working with the cube root.

Use the hint given.

Let [itex]a=\sqrt[3]{x+h}\,,[/itex] and [itex]b=\sqrt[3]{x}\ .[/itex]
 
  • #3
CallMeShady
45
1
I did that but it ended up getting to this expression after some simplifications:
[ (x + h)2/3(x)1/3 - (x + h)2/3(x)1/3 ] / h
as h → 0.

You can already see what the problem is here. The numerator results in 0, which is subsequently divided by h.
 
  • #4
SammyS
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Gold Member
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I did that but it ended up getting to this expression after some simplifications:
[ (x + h)2/3(x)1/3 - (x + h)2/3(x)1/3 ] / h
as h → 0.

You can already see what the problem is here. The numerator results in 0, which is subsequently divided by h.
So, what did you multiply by ?

You want to multiply [itex]\sqrt[3]{x+h}\,-\,\sqrt[3]{x}[/itex] by something that results in cubing each of those terms. Right?
 
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