Question

for the function u=t^3/2 show that the derivative does not exist at t=0 but that the right-hand derivative exist at that point

Answer 1

yes

Answer 2

Are you expected to use the limit definition for this?
Or can you use the friendly rules of differentiation?

Answer 3

im expected to use the limit of definition

Answer 4

\[\large u(t)=t^{3/2}\]

Using the Limit Definition of the Derivative:\[\large u'(t)=\lim_{h \to 0}\frac{(t+h)^{3/2}-t^{3/2}}{h}\]

And from here, let's seeee.

Answer 5

Multiplying the top and bottom by the `conjugate of the top` gives us,\[\large u'(t)=\lim_{h \to 0}\frac{(t+h)^{3/2}-t^{3/2}}{h}\left(\frac{(t+h)^{3/2}+t^{3/2}}{(t+h)^{3/2}+t^{3/2}}\right)\]

\[\large u'(t)=\lim_{h \to 0}\frac{(t+h)^3-t^3}{h\left[(t+h)^{3/2}+t^{3/2}\right]}\]

Answer 6

Understand that step? It's a little tricky :o

Answer 7

Err hmm, maybe we don't need to go into all of that.. maybe we can simply do this,

Answer 8

no, i think we're on the right track..

Answer 9

the multiplication of the conjugate top is confusing can you elaborate further please :)

Answer 10

Normally when we multiply conjugates it's a little more straight forward, like this:\[\large (a-b)(a+b) \qquad=\qquad a^2-b^2\]

But in our case we have something like this:\[\large \left(a^{3/2}-b^{3/2}\right)\left(a^{3/2}+b^{3/2}\right) \qquad=\qquad \left(a^{3/2}\right)^2-\left(b^{3/2}\right)^2\]

When we have an exponent with another exponent being applied like this, our rules of exponents tell us to `multiply the exponents`\[\large \frac{3}{2}\cdot 2 \qquad=\qquad 3\]So we're ending up with,\[\large a^3-b^3\]

Answer 11

too much? did your head esplode?