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anonymous
 5 years ago
A normal line is a line perpendicular to a tangent line at the point where the tangent meets the function. Give the equation of the normal line to the graph of y=3xsqrt{x^2+6}+4 at the point (0,4)
anonymous
 5 years ago
A normal line is a line perpendicular to a tangent line at the point where the tangent meets the function. Give the equation of the normal line to the graph of y=3xsqrt{x^2+6}+4 at the point (0,4)

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TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1so we're gonna need the calculus version of pointslope form yy1=f'(x1)(xx1) which means we need to find dy/dx (at x=x1) then we need the fact that the normal line to this line has the slope of the negative reciprocal 1/(f'(x1) between those facts you're good

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1so step one (as usual in calc) find dy/dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so dy/dx = (3)(1/2(x^2+6)^1/2(2x)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1I hadn't tried yet, but that's not what I'm getting...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did the chain rule for sqrt{x^2+6}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh so would it be 3x/sqrt{x^2+6}

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1Yeah, but you need to use it in conjunction with the product rul so be very careful... I even have messed up a little according to wolfram, so I'm looking where I lost a 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh would the 3x be part of sqrt{x^2+6} in the beginning

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would i have to use the chain rule for (3xsqrt{x^2+6}) or just sqrt{x^2+6}

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1product then chain rule (3x)'(sqrt(x^2+6)+3x(sqrt(x^2+6))' ^^^^^^^^ chain rule here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm i kinda got this for the product rule i used the chain rule in the product rule... 3(sqrt{x^2+6})+1/2(x^2+6)^1/2(2x)(3x)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1yeah, that looks good now the simplification is the annoying bit, I only just found my earlier mistake :/

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1simplify the obvious stuff then factor out (x^2+6)^(1/2) with finess

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dy/dx= 3sqrt{x^2+6}+6x^2/2sqrt{x^2+6}

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1well, you could just use a calculator I suppose... right what you have so far though, just a pain to evaluate

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1so you should keep going and simplify if you can

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dy/dx= 3sqrt{x^2+6}+3x^2/sqrt{x^2+6}

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1you can factor out 3(x^2+6)^(1/2) if you want to do it to look nice

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1dy/dx=3(x^2+6)^(1/2)+3x^2(x^2+6)^(1/2) =3(x^2+6)^(1/2)(x^2+6+x^2)=3(2x^2+6)/sqrt(x^2+6) =6(x^2+3)/sqrt{x^2+6} watch the exponents carefully if that lost you

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1I have to go eat.... Evaluate that at x=0 to get the derivative at that point, which is your slope of the tangent f'(0) the slope perpendicular to that will be the normal slope, so that is the one you want. We know it will be 1/f'(0) from algebra. so use 1/f'(0) in the the slope m in the pointslope formula for a line yy1=m(xx1) See you later perhaps

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thank you so much! have a great dinner :) thank you.
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