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Study23
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Find an equation to the tangent line to the curve at the given point. (So, I have to find the derivative, but Im stuck  not sure if I am doing this correctly). Function is y=x^(5/2) Point is \(\ (4,32). Please show me step by step! I'm really confused!
 one year ago
 one year ago
Study23 Group Title
Find an equation to the tangent line to the curve at the given point. (So, I have to find the derivative, but Im stuck  not sure if I am doing this correctly). Function is y=x^(5/2) Point is \(\ (4,32). Please show me step by step! I'm really confused!
 one year ago
 one year ago

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Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
you can do it the usual way.... (x^n) ' = nx^(n1) n is 5/2
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Okay, so that is the derivative....?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
I thought it was 5/2(x)^(3/2)
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Where do I proceed from here?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
plug in x=4 to find the slope (m) of the tangent to the curve at that point.. use y= mx +b with the point they gave in order to find b
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Oh Okay. I was trying to plug the derivative in for m, but I guess that isn't correct.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
yeah, you just want the value of the derivative at that particular point...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
not the general expression for the slope at any value of x..
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
So Is this the derivative? 5/2(x)^(3/2)
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks for the help!
 one year ago
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