# Math Help - Beginning Derivatives.

1. ## Beginning Derivatives.

Find an equation for the tangent to the curve at the given point.

y=2(sqrt(x)) point- (1,2)

Ok so I think I can work most of it then I get stuck. I am not sure how you guys use the codes so I will try to type it out the best I can.

lim = ( f(a+h)-f(a) ) / (h)
h->0
= f(1+h)-f(1) / h
= (2 sqrt(1+h) - 2) / h

This is where I have gotten to. substituting in now will give me a 0 in my denominator. So where would I go next?

2. Originally Posted by ur5pointos2slo
Find an equation for the tangent to the curve at the given point.

y=2(sqrt(x)) point- (1,2)

Ok so I think I can work most of it then I get stuck. I am not sure how you guys use the codes so I will try to type it out the best I can.
see the first post here.

lim = ( f(a+h)-f(a) ) / (h)
h->0
= f(1+h)-f(1) / h
= (2 sqrt(1+h) - 2) / h

This is where I have gotten to. substituting in now will give me a 0 in my denominator. So where would I go next?
good. now multiply by the conjugate of the numerator over itself. this is a standard trick. (note that the top becomes the factorized for of the difference of two squares).

so you have $\lim_{h \to 0} \frac {2 \sqrt{1 + h} - 2}h \cdot \frac {2 \sqrt{1 + h} + 2}{2 \sqrt{1 + h} + 2}$

now simplify and continue

3. Originally Posted by Jhevon
see the first post here.

good. now multiply by the conjugate of the numerator over itself. this is a standard trick. (note that the top becomes the factorized for of the difference of two squares).

so you have $\lim_{h \to 0} \frac {2 \sqrt{1 + h} - 2}h \cdot \frac {2 \sqrt{1 + h} + 2}{2 \sqrt{1 + h} + 2}$

now simplify and continue
Thanks for that Ill definately take a look at that posting.

Ok I am just starting but I do remember something about the conjugate from algebra. Would I get..

( 4(1+h)-4) / h(2 sqrt(1+h) + 2) ?

4. Originally Posted by ur5pointos2slo
Thanks for that Ill definately take a look at that posting.

Ok I am just starting but I do remember something about the conjugate from algebra. Would I get..

( 4(1+h)-4) / h(2 sqrt(1+h) + 2) ?
yes

5. ( 4(1+h)-4) / h(2 sqrt(1+h) + 2)

Yes well if i plug in for h now im still going to end up with 0 in my denominator..hmmm I am totally lost

6. Originally Posted by ur5pointos2slo
( 4(1+h)-4) / h(2 sqrt(1+h) + 2)

Yes well if i plug in for h now im still going to end up with 0 in my denominator..hmmm I am totally lost
i told you to "simplify and continue"

surely you can simplify the numerator more. you problem h will cancel. just simplify

7. Originally Posted by Jhevon
i told you to "simplify and continue"

surely you can simplify the numerator more. you problem h will cancel. just simplify
ok let me see..

so 4 (1+h) - 4 /(h(2 sqrt(1+h) + 2) will become

4h/ ( h ( 2 sqrt(1+h) + 2) ) so can I distribute the h through on the bottom to get...

4h / 2h + h sqrt(1+h) + 2h? or would this be the wrong direction to go?

8. Originally Posted by ur5pointos2slo
ok let me see..

so 4 (1+h) - 4 /(h(2 sqrt(1+h) + 2) will become

4h/ ( h ( 2 sqrt(1+h) + 2) ) so can I distribute the h through on the bottom to get...

4h / 2h + h sqrt(1+h) + 2h? or would this be the wrong direction to go?
NO! why distribute the h??

you have 4h in the top and an h in the bottom, cancel the h's.

9. Originally Posted by Jhevon
NO! why distribute the h??

you have 4h in the top and an h in the bottom, cancel the h's.
stupid me.. so ill end up with 4/4 which = 1

which is the slope. so my equation will be y= x+1

10. Originally Posted by ur5pointos2slo
stupid me.. so ill end up with 4/4 which = 1

which is the slope. so my equation will be y= x+1
yup