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Factorising (after differentiating) (1 Viewer)

Smile12345

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I'm not sure how to factorise this after differentiating - hopefully I've done this right.....

Q:Find the exact gradient of the tangent to the curve y=x(2x+5)^1/2 at the point where x=1...

I've differentiated to (2x+5)^1/2 + x(2x+5)^-1/2.... I think the x is wrong?

Thanks for your help in advance... :)
 
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spidercrab

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dont think you need to factories, just substitute x = 1 and simplify
 

Menomaths

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I'm not sure how to factorise this after differentiating - hopefully I've done this right.....

Q:Find the exact gradient of the tangent to the curve y=x(2x+5)^1/2 at the point where x=1...

I've differentiated to (2x+5)^1/2 + x(2x+5)^-1/2.... I think the x is wrong?

Thanks for your help in advance... :)
You've differentiated correctly
 

Menomaths

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Very good... Thanks for confirmation... Now how do I factorise?
You sub in x = 1.
(2x+5)^1/2 + x(2x+5)^-1/2
This is (7^1/2) +1 /7^1/2.
Take L.C.F
8/(7^1/2)
 
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Menomaths

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Thanks.... How did you get 8??? :)
When you add square root 7 + 1/square root 7, you take square root 7 as the common denominator and you add the fractions as normal.
Multiply square root 7's denominator (which is 1) by the common denominator which is square root 7 you get (square root 7)*2 which is 7. Then 7+1 = 8
 

Smile12345

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When you add square root 7 + 1/square root 7, you take square root 7 as the common denominator and you add the fractions as normal.
Multiply square root 7's denominator (which is 1) by the common denominator which is square root 7 you get (square root 7)*2 which is 7. Then 7+1 = 8
Thanks heaps... :) I understand exactly... Really was having one of those moments... :) Please forgive me. :)
 

Menomaths

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It's okay, I have a lot of those moments too.
 

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