Find the yintercept of the tangent line to: .4/ sqrt(3 + x) atby Lo.Lee.Ta. Tags: 4 or, line, sqrt3, tangent, yintercept 

#1
Jan913, 06:12 PM

P: 209

1.Find the yintercept of the tangent line to: y= .4/√(3 + x) at [2.5, .170560573084488]
2. So I thought the first step would be to find the slope of the tangent line. I think we find the slope of the tangent line by taking the derivative. So I am going to use the Quotient Rule to take the derivative. [(√3 + x)(0)  (.4)(.5(x + 3)^1/2) * 1] / √(x + 3)^2 .2√(x + 3) / (x + 3) < This is the slope of the tan line to y= .4/√(3 + x) If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x? So, [.2√(3 + 2.5)]/(2.5 + 3) = .085< This is the slope Now, I thought we could substitute everything into y = mx + b in order to find out what b is. .1705606 = .085(2.5) + b b = .3831 3. But this is NOT the right answer! Please tell me what I am doing wrong! Thank you! :) 



#2
Jan913, 06:16 PM

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#3
Jan913, 06:21 PM

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For calculating the derivative, instead of using the quotient rule, rewrite the equation:
y = 0.4/SQRT(x+3) as y = 0.4 * (x+3)^(1/2) Now, you have an eqn. of the form y = a u^n and you can use the power rule to get the derivative. The Quotient Rule is useful when you are dealing with the ratio of two functions. 



#4
Jan913, 06:24 PM

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Find the yintercept of the tangent line to: .4/ sqrt(3 + x) atI see that you are including brackets around your numerator, so you've been listening. Good! You have a mistake above that might just be a transcription error. At the beginning you have (√3 + x)(0) ... That should be [(√(3 + x)(0) dy/dx = .2(x + 3)^{1/2}/(x + 3) 



#5
Jan913, 06:29 PM

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To expand on what SteamKing said, you should NEVER use the quotient rule if either the numerator or denominator is a constant.
It's not incorrect to use it in this case, but it's more complicated, and the chances are greater of making a mistake. For example, if f(x) = 2/(x^{2} + 1), write this as f(x) = 2(x^{2} + 1)^{1} and use the chain rule to get f'(x) = 2(1)(x_{2} + 1)^{2}(2x). 



#6
Jan913, 07:01 PM

P: 209

When I was working it out on paper, I left out the () on the 1/2 exponent!
Thanks for pointing that out! So the Quotient Rule method is: [.4(1/2(x + 3)^1/2)]/(x + 3) .2/(x + 3)^1/2 * (x + 3) = .2/ (x + 3)^3/2 If I substitute the x = 2.5, then I get: .0155 as the slope. I tried the Product Rule as well to see if I'd get the same thing. So, .4*1/2(x + 3)^3/2 + [(x + 3)^1/2](0) = .2(x +3)^3/2 When substituting x = 2.5, I get the same slope of .0155 When I substitute everything into y = mx + b, I get: .17056 = .0155(2.5) + b b = .209 ...But this is not the right answer... Do you know what could be wrong? Is my derivative still wrong? Thanks! :) 



#7
Jan913, 07:54 PM

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#8
Jan1013, 11:43 AM

P: 209

@Dick  Thank you so much for always helping me! :) You are kind.
I used the extra digits you found, and NOW the computer counts it as correct! Thanks! :D 


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