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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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)

Mathematics
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so we're gonna need the calculus version of point-slope form y-y1=f'(x1)(x-x1) 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
so step one (as usual in calc) find dy/dx
so dy/dx = (3)(1/2(x^2+6)^-1/2(2x)

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or 6x/2(x^2+6)
or 3x/x^2+6
I hadn't tried yet, but that's not what I'm getting...
oh
i did the chain rule for sqrt{x^2+6}
oh so would it be 3x/sqrt{x^2+6}
Yeah, 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
that's closer^
oh would the 3x be part of sqrt{x^2+6} in the beginning
would i have to use the chain rule for (3xsqrt{x^2+6}) or just sqrt{x^2+6}
product then chain rule (3x)'(sqrt(x^2+6)+3x(sqrt(x^2+6))' ^^^^^^^^ chain rule here
hmm 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)
yeah, that looks good now the simplification is the annoying bit, I only just found my earlier mistake :/
simplify the obvious stuff then factor out (x^2+6)^(-1/2) with finess
dy/dx= 3sqrt{x^2+6}+6x^2/2sqrt{x^2+6}
well, you could just use a calculator I suppose... right what you have so far though, just a pain to evaluate
so you should keep going and simplify if you can
dy/dx= 3sqrt{x^2+6}+3x^2/sqrt{x^2+6}
you can factor out 3(x^2+6)^(-1/2) if you want to do it to look nice
hmm how?
dy/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
I 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 point-slope formula for a line y-y1=m(x-x1) See you later perhaps
okay thank you so much! have a great dinner :) thank you.

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